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  • (instead of printing)
Java
1
Print Hello World
Print a literal string on standard output 
System.out.println("Hello World");
Alternative implementation:
out.print("Hello, World!");
2
Print Hello 10 times
Loop to execute some code a constant number of times 
for(int i=0;i<10;i++)
  System.out.println("Hello");
Alternative implementation:
System.out.print("Hello\n".repeat(10));
Alternative implementation:
int i = 0;
while (i++ < 10) out.println("Hello");
Alternative implementation:
for (int i = 0; i++ < 10; out.println("Hello"));
Alternative implementation:
int i = 0;
do out.println("Hello");
while (i++ < 10);
Alternative implementation:
range(0, 10).forEach(x -> out.println("Hello"));
Alternative implementation:
generate(() -> "Hello%n")
    .limit(10)
    .forEach(out::printf);
3
Create a procedure
Like a function which doesn't return any value, thus has only side effects (e.g. Print to standard output) 
void finish(String name){
  System.out.println("My job here is done. Goodbye " + name);
}
Alternative implementation:
private void methodName() {
	System.out.println("Hello, World!");
}
Alternative implementation:
void f() { out.println("abc"); }
Alternative implementation:
interface F { void set(); }
F f = () -> out.println("abc");
f.set();
4
Create a function
Create a function which returns the square of an integer 
int square(int x){
  return x*x;
}
Alternative implementation:
Function<Integer,Integer> squareFunction = x -> x * x;
Alternative implementation:
int square(int a) { return a * a; }
Alternative implementation:
int square(int a) { return (int) pow(a, 2); }
Alternative implementation:
interface F { int get(int a); }
F square = x -> x * x;
Alternative implementation:
BigInteger square(BigInteger a) {
    return a.multiply(a);
}
5
Create a 2D Point data structure
Declare a container type for two floating-point numbers x and y 
class Point{
  double x;
  double y;
}
Alternative implementation:
class Point { double x, y; }
Alternative implementation:
public record Point(double x, double y) { }
6
Iterate over list values
Do something with each item x of the list (or array) items, regardless indexes. 
for (Item x: items) {
    doSomething(x);
}
Alternative implementation:
for(int i=0;i<items.length;i++){
	doSomething( items[i] );
}
Alternative implementation:
items.stream().forEach(item -> doSomething(item));
Alternative implementation:
for (T x : items) {}
Alternative implementation:
T t;
Iterator<T> i = items.iterator();
while (i.hasNext()) {
    t = i.next();
}
Alternative implementation:
T t;
Iterator<T> i = items.listIterator();
while (i.hasNext()) {
    t = i.next();
}
7
Iterate over list indexes and values
Print each index i with its value x from an array-like collection items 
int i, n;
for (i = 0, n = items.length; i < n; ++i)
    out.println(i + " = " + items[i]);
Alternative implementation:
int i, n;
for (i = 0, n = items.size(); i < n; ++i)
    out.println(i + " = " + items.get(i));
Alternative implementation:
int i = 0;
for (T t : items)
    out.println(i++ + " = " + t);
Alternative implementation:
range(0, items.length)
    .forEach(i -> {
        out.println(i + " = " + items[i]);
    });
8
Create a map (associative array)
Create a new map object x, and provide some (key, value) pairs as initial content. 
Map<String,Integer> x = new HashMap<>();
x.put("one", 1);
x.put("two", 2);
Alternative implementation:
final Map<String, Integer> x = new HashMap<String, Integer>() {{
    put("one", 1);
    put("two", 2);
    put("three", 3);
}};
Alternative implementation:
Map<String, Integer> x = of("x", 1, "y", 2);
Alternative implementation:
Entry<String, Integer> a = entry("x", 1),
                       b = entry("y", 2);
Map<String, Integer> x = ofEntries(a, b);
Alternative implementation:
Map<String, Integer> x = new HashMap<>(of("x", 1, "y", 2));
9
Create a Binary Tree data structure
The structure must be recursive because left child and right child are binary trees too. A node has access to children nodes, but not to its parent. 
class BinTree<T extends Comparable<T>>{
   T value;
   BinTree<T> left;
   BinTree<T> right;
}
Alternative implementation:
interface Tree<T>{
    boolean search(T x);
}
class Nil<T extends Comparable<? super T>> implements Tree<T>{
    public boolean search(T x) { return false; }
}
class Node<T extends Comparable<? super T>> implements Tree<T> {
 T item;  Tree<T> left; Tree<T> right;
 public Node(T i, Tree<T> l, Tree<T> r) {
     item=i;  left=l;  right=r; }
 public boolean search(T x) {
     int cmp = x.compareTo(item);
     return cmp==0 || (cmp<0 && left.search(x)) || (cmp>0 && right.search(x));
 }
10
Shuffle a list
Generate a random permutation of the elements of list x 
Collections.shuffle(x);
Alternative implementation:
shuffle(asList(x));
11
Pick a random element from a list
The list x must be non-empty. 
x.get((int)(Math.random()*x.size()))
Alternative implementation:
x.get(ThreadLocalRandom.current().nextInt(0, x.size()))
Alternative implementation:
if (!x.isEmpty()) {
    Random r = new Random();
    T t = x.get(r.nextInt(x.size()));
}
12
Check if list contains a value
Check if the list contains the value x.
list is an iterable finite container. 
boolean contains(int[] list, int x){
  for(int y:list)
    if( y==x )
      return true;
  return false;
}
Alternative implementation:
boolean <T> contains(T[] list, T x){
  if( x==null){
    for(T y:list)
      if( y==null )
        return true;
  }else{
    for(T y:list)
      if( x.equals(y) )
        return true;
  }
  return false;
}
Alternative implementation:
list.contains(x)
Alternative implementation:
asList(list).contains(x);
13
Iterate over map keys and values
Access each key k with its value x from an associative array mymap, and print them. 
for (Map.Entry<Object, Object> entry : mymap.entrySet()) {
    Object k = entry.getKey();
    Object x = entry.getValue();
    System.out.println("Key=" + k + ", Value=" + x);
}
Alternative implementation:
mymap.forEach((k,x) -> System.out.println("Key=" + k + ", Value=" + x));
Alternative implementation:
K k;
X x;
for (Entry<K, X> e : mymap.entrySet()) {
    k = e.getKey();
    x = e.getValue();
    out.println(e);
}
14
Pick uniformly a random floating point number in [a..b)
Pick a random number greater than or equals to a, strictly inferior to b. Precondition : a < b. 
double pick(double a, double b){
	return a + (Math.random() * (b-a));
}
Alternative implementation:
Random r = new Random();
double v = r.nextDouble(a, b);
Alternative implementation:
Random r = new Random();
double v = (r.nextDouble() * (b - a)) + a;
15
Pick uniformly a random integer in [a..b]
Pick a random integer greater than or equals to a, inferior or equals to b. Precondition : a < b. 
int pick(int a, int b){
	return a + new Random().nextInt(b - a + 1);
}
Alternative implementation:
int v = new Random().nextInt(a, b + 1);
16
Depth-first traversal of a binary tree
Call a function f on every node of binary tree bt, in depth-first infix order 
void dfs(BinTree bt) {
	if (bt.left != null) {
		dfs(bt.left);
        }
	f(bt);
	if (bt.right != null) {
		dfs(bt.right);
        }
}
Alternative implementation:
class BinTree {
	// ...

	void dfs() {
		if( left != null )
			left.dfs();
		f(this);
		if( right != null )
			right.dfs();
	}
}
Alternative implementation:
class BinTree<T> {
	// ...

	void dfs(Consumer<BinTree<T>> f) {
		if( left != null )
			left.dfs(f);
		f.accept(this);
		if( right != null )
			right.dfs(f);
	}
}
17
Create a Tree data structure
The structure must be recursive. A node may have zero or more children. A node has access to its children nodes, but not to its parent. 
class Tree<K,V> {
  K key;
  V deco;
  List<Tree<K,V>> children = new ArrayList<>();
}
19
Reverse a list
Reverse the order of the elements of the list x.
This may reverse "in-place" and destroy the original ordering. 
static <T> void reverse(List<T> x){
	int n = x.size();
	for(int i=0;i<n/2;i++){
		T tmp = x.get(i);
		x.set(i, x.get(n-i-1));
		x.set(n-i-1, tmp);
	}
}
Alternative implementation:
Collections.reverse(x);
Alternative implementation:
Collections.reverse(x);
Alternative implementation:
int i, m = x.size(), n = m-- / 2;
for (i = 0; i < n; ++i) swap(x, i, m - i);
20
Return two values
Implement a function search which looks for item x in a 2D matrix m.
Return indices i, j of the matching cell.
Think of the most idiomatic way in the language to return the two values at the same time. 
static class Position{
	int i;
	int j;
}

Position search(int[][] m, int x){
	for(int i=0;i<m.length;i++)
		for(int j=0;j<m[i].length;j++)
			if(m[i][j] == x){
				Position pos= new Position();
				pos.i = i;
				pos.j = j;
				return pos;
			}
	return null;
}
Alternative implementation:
record Cell(int i, int j) {}
<T> Cell search(T x, T m[][]) {
    int i, j, M = m.length, N;
    for (i = 0; i < M; ++i) {
        N = m[i].length;
        for (j = 0; j < N; ++j)
            if (m[i][j] == x)
                return new Cell(i, j);
    }
    return null;
}
21
Swap values
Swap the values of the variables a and b
 
T tmp = a;
a = b;
b = tmp;
22
Convert string to integer
Extract the integer value i from its string representation s (in radix 10) 
int i = Integer.parseInt(s);
Alternative implementation:
int i = new Integer(s).intValue();
Alternative implementation:
Integer i = Integer.valueOf(s, 10);
Alternative implementation:
int i = s.chars()
     .map(x -> x - '0')
     .reduce((a, b) -> (a * 10) + b)
     .getAsInt();
23
Convert real number to string with 2 decimal places
Given a real number x, create its string representation s with 2 decimal digits following the dot. 
String s = String.format("%.2f", x);
Alternative implementation:
String s = "%.2f".formatted(x);
Alternative implementation:
MathContext m = new MathContext(3, HALF_UP);
BigDecimal d = new BigDecimal(x, m);
String s = d.toPlainString();
Alternative implementation:
BigDecimal d = new BigDecimal(x);
String s = "%.2f".formatted(d);
Alternative implementation:
NumberFormat f = getNumberInstance();
f.setRoundingMode(HALF_UP);
f.setMaximumFractionDigits(2);
String s = f.format(x);
24
Assign to string the japanese word ネコ
Declare a new string s and initialize it with the literal value "ネコ" (which means "cat" in japanese) 
String s = "ネコ";
25
Send a value to another thread
Share the string value "Alan" with an existing running process which will then display "Hello, Alan" 
https://gist.github.com/f7c174c53efaba4d8575
Alternative implementation:
class Process {
    volatile String s;
    void f() throws InterruptedException {
        Thread a = new Thread(() -> {
            try {
                sleep(1_000);
                out.println("Hello, " + s);
            } catch (InterruptedException e) {
                throw new RuntimeException(e);
            }
        });
        a.start();
        sleep(500);
        s = "Alan";
    }
}
Process p = new Process();
p.f();
26
Create a 2-dimensional array
Declare and initialize a matrix x having m rows and n columns, containing real numbers. 
double[][] x = new double[m][n];
Alternative implementation:
BigDecimal x[][] = new BigDecimal[m][n];
27
Create a 3-dimensional array
Declare and initialize a 3D array x, having dimensions boundaries m, n, p, and containing real numbers. 
double[][][] x = new double[m][n][p];
Alternative implementation:
BigDecimal x[][][] = new BigDecimal[m][n][p];
28
Sort by a property
Sort the elements of the list (or array-like collection) items in ascending order of x.p, where p is a field of the type Item of the objects in items. 
Arrays.sort(items, new Comparator<Item>(){
	public int compare(Item a, Item b){
		return a.p - b.p;
	}
});
Alternative implementation:
Collections.sort(items, new Comparator<Item>(){
	@Override
	public int compare(Item a, Item b){
		return a.p - b.p;
	}
});
Alternative implementation:
items.stream().sorted(Comparator.comparing(x -> x.p))
Alternative implementation:
sort(items, comparing(x -> x.p));
Alternative implementation:
items.sort(comparing(a -> a.p));
29
Remove item from list, by its index
Remove i-th item from list items.
This will alter the original list or return a new list, depending on which is more idiomatic.
Note that in most languages, the smallest valid value for i is 0. 
items.remove(i);
Alternative implementation:
items.remove(i);
30
Parallelize execution of 1000 independent tasks
Launch the concurrent execution of procedure f with parameter i from 1 to 1000.
Tasks are independent and f(i) doesn't return any value.
Tasks need not run all at the same time, so you may use a pool. 
final ExecutorService executor = Executors.newFixedThreadPool(NB_THREADS);
for (int i = 1; i <= 1000; i++) {
  executor.submit(() -> f(i));
}
executor.shutdown();
31
Recursive factorial (simple)
Create the recursive function f which returns the factorial of the non-negative integer i, calculated from f(i-1) 
int f(int i) {
    if (i == 0)
        return 1;
    else
        return i * f(i - 1);
}
32
Integer exponentiation by squaring
Create function exp which calculates (fast) the value x power n.
x and n are non-negative integers. 
int exp(int x, int n){
	if(n==0)
		return 1;
	if(n==1)
		return x;
	if(n%2==0)
		return exp(x*x, n/2);
	else
		return x * exp(x*x, (n-1)/2);
}
33
Atomically read and update variable
Assign to the variable x the new value f(x), making sure that no other thread may modify x between the read and the write. 
synchronized(lock){
  x = f(x);
}
Alternative implementation:
volatile T x;
Alternative implementation:
class Example<T> {
    T x;
    Object lock = new Object();
    T read() {
        synchronized (lock) {
            return x;
        }
    }
    void write(T x) {
        synchronized (lock) {
            this.x = x;
        }
    }
}
34
Create a set of objects
Declare and initialize a set x containing unique objects of type T. 
Set<T> x = new HashSet<T>();
Alternative implementation:
Set<T> x = new HashSet<T>();
x.add(a);
x.add(b);
Alternative implementation:
Set<T> x = of(a, b, c);
35
First-class function : compose
Implement a function compose (A -> C) with parameters f (A -> B) and g (B -> C), which returns the composition function g ∘ f 
public Function<A, C> compose(Function<A, B> f, Function<B, C> g) {
   return x -> g.apply(f.apply(x));
}
Alternative implementation:
Function<Integer, Integer> addOne = i-> i + 1;
Function<Integer, String> toString = i-> i.toString();
Function<Integer, String> printIncremented = toString.compose(addOne);
Alternative implementation:
<A, B, C> Function<A, C> compose(Function<A, B> f, Function<B, C> g) {
    return f.andThen(g);
}
36
First-class function : generic composition
Implement a function compose which returns composition function g ∘ f for any functions f and g having exactly 1 parameter. 
    default <V> Function<V, R> compose(Function<? super V, ? extends T> before) {
        Objects.requireNonNull(before);
        return (V v) -> apply(before.apply(v));
    }
37
Currying
Transform a function that takes multiple arguments into a function for which some of the arguments are preset. 
IntBinaryOperator simpleAdd = (a, b) -> a + b;
IntFunction<IntUnaryOperator> curriedAdd = a -> b -> a + b;
System.out.println(simpleAdd.applyAsInt(4, 5));
System.out.println(curriedAdd.apply(4).applyAsInt(5));
38
Extract a substring
Find substring t consisting in characters i (included) to j (excluded) of string s.
Character indices start at 0 unless specified otherwise.
Make sure that multibyte characters are properly handled. 
String t = s.substring(i,j);
39
Check if string contains a word
Set the boolean ok to true if the string word is contained in string s as a substring, or to false otherwise. 
boolean ok = s.contains(word);
Alternative implementation:
boolean ok = s.indexOf(word) != -1;
40
Graph with adjacency lists
Declare a Graph data structure in which each Vertex has a collection of its neighbouring vertices. 
class Graph{
  List<Vertex> vertices;

  static class Vertex{
    int id;
    List<Vertex> neighbours;
  }
}
Alternative implementation:
class Graph{
  Set<Vertex> vertices;

  static class Vertex{
    int id;
    Set<Vertex> neighbours;
  }
}
41
Reverse a string
Create the string t containing the same characters as the string s, in reverse order.
The original string s must remain unaltered. Each character must be handled correctly regardless its number of bytes in memory. 
String t = new StringBuilder(s).reverse().toString();
Alternative implementation:
String t = "";
for (char c : s.toCharArray())
    t = c + t;
Alternative implementation:
char a[] = s.toCharArray(), c;
int i, m = a.length, n = m-- / 2, z;
for (i = 0; i < n; ++i) {
    c = a[i];
    a[i] = a[z = m - i];
    a[z] = c;
}
String t = new String(a);
Alternative implementation:
String t = s.chars()
    .mapToObj(x -> valueOf((char) x))
    .reduce((a, b) -> b + a)
    .get();
42
Continue outer loop
Print each item v of list a which is not contained in list b.
For this, write an outer loop to iterate on a and an inner loop to iterate on b. 
mainloop: for(int v:a){
	for(int w:b){
		if(v==w)
			continue mainloop;
	}
	System.out.println(v);
}
43
Break outer loop
Look for a negative value v in 2D integer matrix m. Print it and stop searching. 
mainloop: for(int i=0;i<m.length;i++)
	for(int j=0;j<m[i].length;j++)
		if(m[i][j]<0){
			System.out.println(m[i][j]);
			break mainloop;
		}
Alternative implementation:
   int i, j, M = m.length, N = m[0].length, x;
a: for (i = 0; i < M; ++i)
       for (j = 0; j < N; ++j)
           if ((x = m[i][j]) < 0) {
               out.println(x);
               break a;
           }
Alternative implementation:
int i, j, M = m.length, N = m[0].length, x;
boolean b = false;
for (i = 0; i < M; ++i) {
    for (j = 0; j < N; ++j)
        if ((x = m[i][j]) < 0) {
            out.println(x);
            b = true;
            break;
        }
    if (b) break;
}
44
Insert element in list
Insert the element x at position i in the list s. Further elements must be shifted to the right. 
s.add(i, x);
Alternative implementation:
<T> T[] insert(T[] s, T x, int i) {
    int a = 0, n = s.length;
    Class<?> c = s.getClass().getComponentType();
    T t[] = (T[]) newInstance(c, n + 1);
    arraycopy(s, a, t, a, i);
    t[i] = x;
    arraycopy(s, i, t, i + 1, n - i);
    return t;
}
45
Pause execution for 5 seconds
Sleep for 5 seconds in current thread, before proceeding with the next instructions. 
Thread.sleep(5000);
Alternative implementation:
TimeUnit.SECONDS.sleep(5);
46
Extract beginning of string (prefix)
Create the string t consisting of the 5 first characters of the string s.
Make sure that multibyte characters are properly handled. 
String t = s.substring(0,5);
47
Extract string suffix
Create string t consisting in the 5 last characters of string s.
Make sure that multibyte characters are properly handled. 
String t = s;
if (s.length()>= 5)
	t = s.substring(s.length()-5);
Alternative implementation:
int i = s.length() - 5;
if (isSurrogate(s.charAt(i))) --i;
String t = s.substring(i);
48
Multi-line string literal
Assign to variable s a string literal consisting in several lines of text, including newlines. 
String s = "This is a very long string which needs \n" +
           "to wrap across multiple lines because \n" +
           "otherwise my code is unreadable.";
Alternative implementation:
String s = """
This is a very long string which needs
to wrap across multiple lines because
otherwise my code is unreadable.
""";
Alternative implementation:
String s, n = lineSeparator();
s = "line 1" + n;
s = s + "line 2" + n;
s = s + "line 3" + n;
s = s + "line 4";
Alternative implementation:
String s;
StringBuilder b = new StringBuilder();
Formatter f = new Formatter(b);
f.format("line 1%n");
f.format("line 2%n");
f.format("line 3%n");
f.format("line 4");
f.flush();
s = b.toString();
49
Split a space-separated string
Build list chunks consisting in substrings of the string s, separated by one or more space characters. 
String chunks[] = s.split(" +");
Alternative implementation:
List<String> chunks = of(s.split(" +"));
Alternative implementation:
List<String> chunks = asList(s.split(" +"));
Alternative implementation:
Pattern p = compile(" +");
Scanner t = new Scanner(s).useDelimiter(p);
List<String> chunks = t.tokens().toList();
t.close();
Alternative implementation:
Pattern p = compile(" +");
Scanner t = new Scanner(s).useDelimiter(p);
String chunks[] = t.tokens().toArray(String[]::new);
t.close();
Alternative implementation:
List<String> chunks
    = new ArrayList<>(of(s.split(" +")));
50
Make an infinite loop
Write a loop that has no end clause. 
for(;;){
	// Do something
}
Alternative implementation:
while(true) {
	// Do something	
}
Alternative implementation:
do {} while (true);
Alternative implementation:
while (true);
Alternative implementation:
for (;;);
51
Check if map contains key
Determine whether the map m contains an entry for the key k 
m.containsKey(k)
52
Check if map contains value
Determine whether the map m contains an entry with the value v, for some key. 
m.containsValue(v)
53
Join a list of strings
Concatenate elements of string list x joined by the separator ", " to create a single string y. 
String y = String.join(", ", x);
Alternative implementation:
String y = x.get(0);
for (int i = 1, n = x.size(); i < n; ++i)
    y = y + ", " + x.get(i);
Alternative implementation:
String y = x.stream()
    .reduce((a, b) -> a + ", " + b)
    .get();
54
Compute sum of integers
Calculate the sum s of the integer list or array x. 
int s = 0;
for (int i : x) {
  s += i;
}
Alternative implementation:
BigInteger s = new BigInteger(valueOf(x[0]));
for (int i = 1, n = x.length; i < n; ++i)
    s = s.add(new BigInteger(valueOf(x[i])));
Alternative implementation:
BigInteger s = stream(x)
    .mapToObj(String::valueOf)
    .map(BigInteger::new)
    .reduce(BigInteger::add)
    .get();
Alternative implementation:
int i = 1, n = x.length, s = x[0];
while (i < n) s = s + x[i++];
Alternative implementation:
int s = of(x).sum();
55
Convert integer to string
Create the string representation s (in radix 10) of the integer value i. 
String s=((Integer)i).toString();
Alternative implementation:
String s = Integer.toString(i);
Alternative implementation:
String s = String.valueOf(i);
Alternative implementation:
String s = "" + i;
Alternative implementation:
String s = "%d".formatted(i);
56
Launch 1000 parallel tasks and wait for completion
Fork-join : launch the concurrent execution of procedure f with parameter i from 1 to 1000.
Tasks are independent and f(i) doesn't return any value.
Tasks need not run all at the same time, so you may use a pool.
Wait for the completion of the 1000 tasks and then print "Finished". 
class Task implements Runnable {
	int i;
	Task(int i) {
		this.i = i;
	}
	@Override
	public void run() {
		f(i);
	}
}

ExecutorService executor = Executors.newFixedThreadPool(4);
for (int i = 1; i <= 1000; i++) {
	Task task = new Task(i);
	executor.submit(task);
}
executor.shutdown();
executor.awaitTermination(10L, TimeUnit.MINUTES);
System.out.println("Finished");
Alternative implementation:
ExecutorService executor = Executors.newFixedThreadPool(4);
for (int i = 1; i <= 1000; i++) {
	Task task = new Task(i);
	executor.submit(() -> f(i));
}
executor.shutdown();
executor.awaitTermination(10L, TimeUnit.MINUTES);
System.out.println("Finished");
57
Filter list
Create the list y containing the items from the list x that satisfy the predicate p. Respect the original ordering. Don't modify x in-place. 
List<T> y = x.stream().filter(p).toList();
Alternative implementation:
List<T> y = new ArrayList<>();
for (T t : x) if (p.test(t)) y.add(t);
58
Extract file content to a string
Create the string lines from the content of the file with filename f. 
byte[] encoded = Files.readAllBytes(Paths.get(f));
String lines = new String(encoded, StandardCharsets.UTF_8);
Alternative implementation:
File F = new File(f);
try (Scanner s = new Scanner(F)) {
    s.useDelimiter("\\R");
    String n = lineSeparator(),
           lines = s.tokens()
                    .collect(joining(n));
}
59
Write to standard error stream
Print the message "x is negative" to standard error (stderr), with integer x value substitution (e.g. "-2 is negative"). 
System.err.format("%d is negative\n",x);
Alternative implementation:
System.err.printf("%d is negative", x);
60
Read command line argument
Assign to x the string value of the first command line parameter, after the program name. 
String x = args[0];
Alternative implementation:
public static void main(String[] args) {
    if (args.length != 0) {
        String x = args[0];
    }
}
61
Get current date
Assign to the variable d the current date/time value, in the most standard type. 
Instant d = Instant.now();
Alternative implementation:
Date a = getInstance().getTime();
String d = "%tc".formatted(a);
Alternative implementation:
long a = currentTimeMillis();
String d = "%tc".formatted(a);
62
Find substring position
Set i to the first position of string y inside string x, if exists.

Specify if i should be regarded as a character index or as a byte index.

Explain the behavior when y is not contained in x. 
int i = x.indexOf(y);
63
Replace fragment of a string
Assign to x2 the value of string x with all occurrences of y replaced by z.
Assume occurrences of y are not overlapping. 
String x2 = x.replace(y, z);
64
Big integer : value 3 power 247
Assign to x the value 3^247 
BigInteger x = new BigInteger("3", 10).pow(247);
Alternative implementation:
BigInteger x = new BigInteger("3").pow(247);
65
Format decimal number
From the real value x in [0,1], create its percentage string representation s with one digit after decimal point. E.g. 0.15625 -> "15.6%" 
String s = new DecimalFormat("0.0%").format(x);
Alternative implementation:
String s = "%.1f%%".formatted(x * 100);
Alternative implementation:
NumberFormat f = getPercentInstance();
f.setMaximumFractionDigits(1);
f.setRoundingMode(HALF_UP);
String s = f.format(x);
66
Big integer exponentiation
Calculate the result z of x power n, where x is a big integer and n is a positive integer. 
BigInteger z = x.pow(n);
67
Binomial coefficient "n choose k"
Calculate binom(n, k) = n! / (k! * (n-k)!). Use an integer type able to handle huge numbers. 
static BigInteger binom(int N, int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}
68
Create a bitset
Create an object x to store n bits (n being potentially large). 
BitSet x = new BitSet(n);
69
Seed random generator
Use seed s to initialize a random generator.

If s is constant, the generator output will be the same each time the program runs. If s is based on the current value of the system clock, the generator output will be different each time. 
Random r = new Random(s);
70
Use clock as random generator seed
Get the current datetime and provide it as a seed to a random generator. The generator sequence will be different at each run. 
Random rand = new Random(System.currentTimeMillis());
71
Echo program implementation
Basic implementation of the Echo program: Print all arguments except the program name, separated by space, followed by newline.
The idiom demonstrates how to skip the first argument if necessary, concatenate arguments as strings, append newline and print it to stdout. 
public class Echo {
    public static void main(final String... args) {
        out.println(join(" ", args));
    }
}
Alternative implementation:
class Echo {
    public static void main(String[] args) {
        out.print(join(" ", args));
    }
}
74
Compute GCD
Compute the greatest common divisor x of big integers a and b. Use an integer type able to handle huge numbers. 
BigInteger x = a.gcd(b);
75
Compute LCM
Compute the least common multiple x of big integers a and b. Use an integer type able to handle huge numbers. 
BigInteger a = new BigInteger("123456789");
BigInteger b = new BigInteger("987654321");
BigInteger x = a.multiply(b).divide(a.gcd(b));
76
Binary digits from an integer
Create the string s of integer x written in base 2.

E.g. 13 -> "1101" 
String s = Integer.toBinaryString(x);
77
Complex number
Declare a complex x and initialize it with value (3i - 2). Then multiply it by i. 
Complex x = new Complex(-2.0, 3.0);
x = x.multiply(Complex.I);
78
"do while" loop
Execute a block once, then execute it again as long as boolean condition c is true. 
do {
	someThing();
	someOtherThing();
} while(c);
Alternative implementation:
do f();
while (c);
79
Convert integer to floating point number
Declare the floating point number y and initialize it with the value of the integer x . 
int x = 10
float y = (float)x;
System.out.println(y);
Alternative implementation:
Float y = Float.parseFloat(String.valueOf(x));
Alternative implementation:
float y = x;
Alternative implementation:
float y = new BigDecimal(x).floatValue();
80
Truncate floating point number to integer
Declare integer y and initialize it with the value of floating point number x . Ignore non-integer digits of x .
Make sure to truncate towards zero: a negative x must yield the closest greater integer (not lesser). 
int y = (int)x;
81
Round floating point number to integer
Declare the integer y and initialize it with the rounded value of the floating point number x .
Ties (when the fractional part of x is exactly .5) must be rounded up (to positive infinity). 
long y = Math.round(x);
Alternative implementation:
NumberFormat f = getNumberInstance();
f.setRoundingMode(HALF_UP);
f.setMaximumFractionDigits(0);
int y = parseInt(f.format(x));
82
Count substring occurrences
Find how many times string s contains substring t.
Specify if overlapping occurrences are counted. 
int count = StringUtils.countMatches(s, t);
Alternative implementation:
Pattern pattern = Pattern.compile(Pattern.quote(t));
Matcher matcher = pattern.matcher(s);
int count = 0;
while(matcher.find()) count++;
Alternative implementation:
int z = 0;
Matcher m = compile(quote(t)).matcher(s);
while (m.find()) ++z;
Alternative implementation:
int z, i = z = 0;
Matcher m = compile(quote(t)).matcher(s);
while (m.find(i)) {
    ++z;
    i = m.start() + 1;
}
83
Regex with character repetition
Declare regular expression r matching the strings "http", "htttp", "httttp", etc. 
Pattern r = Pattern.compile("htt+p");
Alternative implementation:
Pattern r = compile("ht{2,}p");
84
Count bits set in integer binary representation
Count number c of 1s in the integer i in base 2.

E.g. i=6 → c=2 
int c = Integer.bitCount(i);
Alternative implementation:
int c = new BigInteger(valueOf(i), 2).bitCount();
85
Check if integer addition will overflow
Write boolean function addingWillOverflow which takes two integers x, y and return true if (x+y) overflows.

An overflow may be above the max positive value, or below the min negative value. 
public boolean addingWillOverflow(int x, int y){
    boolean willOverFlow = false;
    if(x > 0 && y > 0){
        if(y > (Integer.MAX_VALUE - x)){
            willOverFlow = true;
        }
    }
    if(x < 0 && y < 0){
       if(y < (Integer.MIN_VALUE - x)){
           willOverFlow = true;
       }
    }
    return willOverFlow;
}
Alternative implementation:
boolean willOverflow(int x, int y) {
    try {
        addExact(x, y);
        subtractExact(x, y);
        return false;
    } catch (ArithmeticException e) {
        return true;
    }
}
86
Check if integer multiplication will overflow
Write the boolean function multiplyWillOverflow which takes two integers x, y and returns true if (x*y) overflows.

An overflow may reach above the max positive value, or below the min negative value. 
static boolean multiplyWillOverflow(int x, int y) {
	return Integer.MAX_VALUE/x < y;
}
Alternative implementation:
boolean b = Math.multiplyExact(x, y);
87
Stop program
Exit immediately.
If some extra cleanup work is executed by the program runtime (not by the OS itself), describe it. 
System.exit(0);
Alternative implementation:
throw new RuntimeException();
Alternative implementation:
throw new Error();
88
Allocate 1M bytes
Create a new bytes buffer buf of size 1,000,000. 
byte[] buf = new byte[1000000];
Alternative implementation:
byte buf[] = new byte[1_000_000];
Alternative implementation:
byte buf[] = new byte[(int) 1e6];
89
Handle invalid argument
You've detected that the integer value of argument x passed to the current function is invalid. Write the idiomatic way to abort the function execution and signal the problem. 
throw new IllegalArgumentException("Invalid value for x:" + x);
Alternative implementation:
<T> void f(T x) {
    Objects.requireNonNull(x);
}
Alternative implementation:
<T> void f(T x) throws Exception {
    if (x != value) throw new Exception();
}
Alternative implementation:
void f(int x) {
    Objects.checkIndex(x, 123);
}
90
Read-only outside
Expose a read-only integer x to the outside world while being writable inside a structure or a class Foo. 
public class Foo {
    private int x;

    public int getX() {
        return x;
    }
}
Alternative implementation:
public class Foo {
    private int x;
    private void set(int x) { this.x = x; }
    public int get() { return x; }
}
93
Pass a runnable procedure as parameter
Implement the procedure control which receives one parameter f, and runs f. 
static void control(Runnable f) {
    f.run();
}
94
Print the type of a variable
Print the name of the type of x. Explain if it is a static type or dynamic type.

This may not make sense in all languages. 
System.out.println(((Object)x).getClass().getName());
Alternative implementation:
var x = "abc";
out.println(x.getClass().getSimpleName());
95
Get file size
Assign to variable x the length (number of bytes) of the local file at path. 
long x = new File(path).length();
96
Check string prefix
Set the boolean b to true if string s starts with prefix prefix, false otherwise. 
boolean b = s.startsWith(prefix);
97
Check string suffix
Set boolean b to true if string s ends with string suffix, false otherwise. 
boolean b = s.endsWith(suffix);
98
Epoch seconds to date object
Convert a timestamp ts (number of seconds in epoch-time) to a date with time d. E.g. 0 -> 1970-01-01 00:00:00 
Date d = new Date(ts * 1000);
Alternative implementation:
Instant d = Instant.ofEpochSecond(ts);
Alternative implementation:
LocalDateTime d = LocalDateTime.ofEpochSecond(ts, 0, ZoneOffset.UTC);
Alternative implementation:
long ts = currentTimeMillis();
String d = "%tc".formatted(ts);
Alternative implementation:
Date ts = getInstance().getTime();
String d = "%s".formatted(ts);
99
Format date YYYY-MM-DD
Assign to the string x the value of the fields (year, month, day) of the date d, in format YYYY-MM-DD. 
String x = new SimpleDateFormat("yyyy-MM-dd").format(d);
Alternative implementation:
String x = String.format("%1$tY-%1$tm-%1$td", d)
Alternative implementation:
Date d = getInstance().getTime();
String s = "%tY-%<tm-%<td".formatted(d);
100
Sort by a comparator
Sort elements of array-like collection items, using a comparator c. 
Arrays.sort(items, c);
Alternative implementation:
Collections.sort(items, c);
101
Load from HTTP GET request into a string
Make an HTTP request with method GET to the URL u, then store the body of the response in the string s. 
String s = HttpClient.newHttpClient().send(HttpRequest.newBuilder()
                .uri(URI.create(u))
                .GET()
                .build(), HttpResponse.BodyHandlers.ofString())
                .body();
Alternative implementation:
HttpsURLConnection h = null;
try {
    h = (HttpsURLConnection) u.openConnection();
    h.setRequestMethod("GET");
    String s = new String(h.getInputStream().readAllBytes());
} catch (IOException e) {
    throw new RuntimeException(e);
} finally {
    if (h != null) h.disconnect();
}
105
Current executable name
1 
String s = System.getProperty("sun.java.command");
106
Get program working directory
Assign to string dir the path of the working directory.
(This is not necessarily the folder containing the executable itself) 
String path = this.getClass().getClassLoader().getResource("").getPath();
Alternative implementation:
String dir = new File("").getAbsolutePath();
Alternative implementation:
String dir = System.getProperty("user.dir");
108
Determine if variable name is defined
Print the value of variable x, but only if x has been declared in this program.
This makes sense in some languages, not all of them. (Null values are not the point, rather the very existence of the variable.) 
try {
    Field f = getClass().getDeclaredField("x");
    out.println(f.get(this));
} catch (NoSuchFieldException e) {

} catch (IllegalAccessException e) {

}
109
Number of bytes of a type
Set n to the number of bytes of a variable t (of type T). 
<T> int bytes(T t) {
    return switch (t.getClass().getSimpleName()) {
        case "Boolean" -> 1;
        case "Byte" -> Byte.BYTES;
        case "Short" -> Short.BYTES;
        case "Character" -> Character.BYTES;
        case "Integer" -> Integer.BYTES;
        case "Float" -> Float.BYTES;
        case "Long" -> Long.BYTES;
        case "Double" -> Double.BYTES;
        default -> -1;
    };
}
110
Check if string is blank
Set the boolean blank to true if the string s is empty, or null, or contains only whitespace ; false otherwise. 
boolean blank = StringUtils.isBlank(s);
Alternative implementation:
boolean blank = s==null || s.strip().isEmpty();
Alternative implementation:
boolean blank = s==null || s.isBlank();
Alternative implementation:
boolean blank = s == null || s.isBlank();
111
Launch other program
From current process, run program x with command-line parameters "a", "b". 
Runtime.getRuntime().exec(new String[]{"x", "a", "b"});
Alternative implementation:
var b = new ProcessBuilder(of("x", "a", "b"));
try {
    b.start().waitFor();
} catch (Exception e) {
    throw new RuntimeException(e);
}
112
Iterate over map entries, ordered by keys
Print each key k with its value x from an associative array mymap, in ascending order of k. 
SortedMap<K, V> mymap = new TreeMap<>();
...
for(Map.Entry<K, V> e: mymap.entrySet())
	System.out.println("Key=" + e.getKey() + ", Value=" + e.getValue());
Alternative implementation:
List<K> keys = new ArrayList<>(mymap.keySet());
Collections.sort(keys);
for(K k: keys)
	System.out.println("Key=" + k + ", Value=" + mymap.get(k));
Alternative implementation:
var map = Map.of("a", 1, "d", 4, "c", 3, "b", 2);
new TreeMap<>(map).entrySet().forEach(System.out::println);
Alternative implementation:
mymap.entrySet().stream().sorted(Entry.comparingByKey()).forEach(System.out::println);
Alternative implementation:
for (Entry<K, V> e : new TreeMap<>(m).entrySet())
    out.println(e);
Alternative implementation:
mymap.entrySet()
    .stream()
    .sorted(comparingByKey())
    .forEach(e -> {
        K k = e.getKey();
        X x = e.getValue();
        out.println(e);
    });
113
Iterate over map entries, ordered by values
Print each key k with its value x from an associative array mymap, in ascending order of x.
Multiple entries may exist for the same value x. 
mymap.entrySet().stream()
    .sorted(Comparator.comparing(Map.Entry::getValue))
    .forEach(entry -> {
        K k = entry.getKey();
        X x = entry.getValue();
        System.out.println("k:" + k + ", x:" + x);
});
Alternative implementation:
mymap.entrySet()
    .stream()
    .sorted(comparingByValue())
    .forEach(out::println);
Alternative implementation:
mymap.entrySet()
    .stream()
    .sorted(comparingByValue())
    .forEach(e -> {
        K k = e.getKey();
        X x = e.getValue();
    });
115
Compare dates
Set boolean b to true if date d1 is strictly before date d2 ; false otherwise. 
boolean before = (d1.compareTo(d2) == -1);
116
Remove occurrences of word from string
Remove all occurrences of string w from string s1, and store the result in s2. 
String s2 = s1.replace(w, "");
117
Get list size
Set n to the number of elements of the list x. 
int n = x.size();
Alternative implementation:
int n = x.length;
118
List to set
Create the set y from the list x.
x may contain duplicates. y is unordered and has no repeated values. 
Set<T> y = new HashSet<>(x);
Alternative implementation:
Set<T> y = new HashSet<T>(x);
Alternative implementation:
Set<T> y = x.stream().collect(toSet());
Alternative implementation:
Set<T> y = copyOf(x);
Alternative implementation:
Set<T> y = new LinkedHashSet<>(x);
119
Deduplicate list
Remove duplicates from the list x.
Explain if the original order is preserved. 
Set<T> uniques = new HashSet<>(x);
x.clear();
x.addAll(uniques);
Alternative implementation:
x = new ArrayList<T>(new HashSet<T>(x));
Alternative implementation:
final HashSet<T> seen = new HashSet<T>();
final Iterator<T> listIt = x.iterator();
while (listIt.hasNext()) {
  final T curr = listIt.next();
  if (seen.contains(curr)) {
    listIt.remove();
  } else {
    seen.add(curr);
  }
}
Alternative implementation:
x = new ArrayList<>(new LinkedHashSet<>(x));
Alternative implementation:
Iterator<T> g;
T t;
int i, n;
for (i = 0, n = x.size(); i < n; ++i) {
    t = x.get(i);
    g = x.listIterator(i + 1);
    while (g.hasNext())
        if (g.next().equals(t)) {
            g.remove();
            --n;
        }
}
120
Read integer from stdin
Read an integer value from the standard input into the variable n 
Scanner in = new Scanner(System.in);
n = in.nextInt();
Alternative implementation:
Scanner in = new Scanner(System.in);
int n = in.nextInt();
Alternative implementation:
Scanner s = new Scanner(System.in);
int n = parseInt(s.nextLine());
s.close();
121
UDP listen and read
Listen UDP traffic on port p and read 1024 bytes into the buffer b. 
int len = 1024;
int p = 8888;
byte[] b = new byte[len];
try (DatagramSocket socket = new DatagramSocket(p)) {
  DatagramPacket packet = new DatagramPacket(b, len);
  socket.receive(packet);
}
122
Declare an enumeration
Create an enumerated type Suit with 4 possible values SPADES, HEARTS, DIAMONDS, CLUBS. 
enum Suit{
  SPADES,
  HEARTS,
  DIAMONDS,
  CLUBS;
}
123
Assert condition
Verify that predicate isConsistent returns true, otherwise report assertion violation.
Explain if the assertion is executed even in production environment or not. 
assert isConsistent() : "State consistency violated";
124
Binary search for a value in sorted array
Write the function binarySearch which returns the index of an element having the value x in the sorted array a, or -1 if no such element exists. 
static int binarySearch(final int[] arr, final int key) {
    final int index = Arrays.binarySearch(arr, key);
    return index < 0 ? - 1 : index;
}
Alternative implementation:
int i = binarySearch(a, x);
125
Measure function call duration
measure the duration t, in nanoseconds, of a call to the function foo. Print this duration. 
long t = System.nanoTime();
foo();
t = System.nanoTime() - t;
System.out.println(t + "ns");
Alternative implementation:
long t = nanoTime();
foo();
out.println(nanoTime() - t);
126
Multiple return values
Write a function foo that returns a string and a boolean value. 
static Object[] returnAnything() {
    return new Object[]{"a string", true};
}

public static void main (String[] args) {
    Object[] array = returnAnything();
    System.out.println(array[0]);
    System.out.println(array[1]);
}
Alternative implementation:
record Tuple(Object ... a) {}
Tuple foo() {
    return new Tuple("abc", true);
}
Alternative implementation:
Object[] foo() {
    return new Object[] { "abc", true };
}
128
Breadth-first traversing of a tree
Call a function f on every node of a tree, in breadth-first prefix order 
static void breadthFirstSearch(Node root, Consumer<Node> f) {
    Queue<Node> queue = new LinkedList<>();
    queue.offer(root);

    while(!queue.isEmpty()) {
        Node polled = queue.poll();
        f.accept(polled);
        polled.children.forEach(a -> queue.offer(a));
    }
}
131
Successive conditions
Execute f1 if condition c1 is true, or else f2 if condition c2 is true, or else f3 if condition c3 is true.
Don't evaluate a condition when a previous condition was true. 
if (c1) {
   f1();
} else if (c2) {
   f2();
} else if (c3) { 
   f3();
}
Alternative implementation:
if (c1) f1();
else if (c2) f2();
else if (c3) f3();
132
Measure duration of procedure execution
Run the procedure f, and return the duration of the execution of f. 
long clock(Runnable f) {
    long t0 = System.currentTimeMillis();
    f.run();
    long t1 = System.currentTimeMillis();
    return t1 - t0;
}
133
Case-insensitive string contains
Set boolean ok to true if string word is contained in string s as a substring, even if the case doesn't match, or to false otherwise. 
ok = s.toLowerCase().contains(word.toLowerCase());
Alternative implementation:
Pattern p = compile("(?i)" + quote(word));
boolean ok = p.matcher(s).find();
134
Create a new list
Declare and initialize a new list items, containing 3 elements a, b, c. 
List<T> items = new ArrayList<>();
items.add(a);
items.add(b);
items.add(c);
Alternative implementation:
List<T> items = Arrays.asList(a, b, c);
Alternative implementation:
var items = List.of(a, b, c);
Alternative implementation:
List<T> items = new ArrayList<>(of(a, b, c));
135
Remove item from list, by its value
Remove at most 1 item from list items, having the value x.
This will alter the original list or return a new list, depending on which is more idiomatic.
If there are several occurrences of x in items, remove only one of them. If x is absent, keep items unchanged. 
items.stream().findFirst().filter(item -> "x".equals(item)).ifPresent(removeIndex -> items.remove(removeIndex));
Alternative implementation:
T value;
for(int index = 0; index < items.size(); index++) {
	value = items.get(index);
	if(value.equals(x)) {
		items.remove(index);
		break;
	}
}
Alternative implementation:
<T> void remove(List<T> items, T x) {
    Iterator<T> i = items.listIterator();
    while (i.hasNext())
        if (i.next() == x) {
            i.remove();
            break;
        }
}
136
Remove all occurrences of a value from a list
Remove all occurrences of the value x from list items.
This will alter the original list or return a new list, depending on which is more idiomatic. 
items.removeAll(Collections.singleton(x));
Alternative implementation:
Iterator<T> i = items.listIterator();
while (i.hasNext())
    if (i.next().equals(x)) i.remove();
Alternative implementation:
items.removeIf(t -> t.equals(x));
137
Check if string contains only digits
Set the boolean b to true if the string s contains only characters in the range '0'..'9', false otherwise. 
boolean b = s.matches("[0-9]*");
Alternative implementation:
boolean b = s.matches("[0-9]+");
Alternative implementation:
boolean b = s.chars()
    .allMatch(Character::isDigit);
if (s.length() == 0) b = false;
Alternative implementation:
boolean b = s.matches("\\d+");
Alternative implementation:
String d = "0123456789";
int i, n = s.length();
boolean b = n != 0;
for (i = 0; i < n; ++i)
    if (d.indexOf(s.charAt(i)) == -1) {
        b = false;
        break;
    }
138
Create temp file
Create a new temporary file on the filesystem. 
File tempFile = File.createTempFile("prefix-", "-suffix");
tempFile.deleteOnExit();
139
Create temp directory
Create a new temporary folder on filesystem, for writing. 
File folder = new File("/path/");
folder.deleteOnExit();
140
Delete map entry
Delete from map m the entry having key k.

Explain what happens if k is not an existing key in m. 
map.remove(k);
141
Iterate in sequence over two lists
Iterate in sequence over the elements of the list items1 then items2. For each iteration print the element. 
items1.forEach(System.out::println);
items2.forEach(System.out::println);
Alternative implementation:
for (E e : items1) out.println(e);
for (E e : items2) out.println(e);
Alternative implementation:
Stream.concat(
	items1.stream(), 
	items2.stream()
).forEach(System.out::println);
142
Hexadecimal digits of an integer
Assign to string s the hexadecimal representation (base 16) of integer x.

E.g. 999 -> "3e7" 
String s = Integer.toHexString(x);
Alternative implementation:
String s = "%x".formatted(x);
Alternative implementation:
String s = HexFormat.of().toHexDigits(x);
143
Iterate alternatively over two lists
Iterate alternatively over the elements of the lists items1 and items2. For each iteration, print the element.

Explain what happens if items1 and items2 have different size. 
Iterator<String> iter1 = items1.iterator();
Iterator<String> iter2 = items2.iterator();
while (iter1.hasNext() || iter2.hasNext()) {
	if (iter1.hasNext()) {
		System.out.println(iter1.next());
	}
	if (iter2.hasNext()) {
		System.out.println(iter2.next());
	}
}
Alternative implementation:
IntStream.range(0, Math.max(items1.size(), items2.size()))
	.boxed()
	.flatMap(idx -> Stream.of(
		items1.size() > idx ? items1.get(idx) : null,
		items2.size() > idx ? items2.get(idx) : null
	))
	.filter(Objects::nonNull)
	.forEach(System.out::println);
Alternative implementation:
int i, A, B, n = max(A = a.size(), B = b.size());
for (i = 0; i < n; ++i) {
    if (i < A) out.println(a.get(i));
    if (i < B) out.println(b.get(i));
}
Alternative implementation:
Iterator<T> A = a.iterator(), B = b.iterator();
boolean x = A.hasNext(), y = B.hasNext();
while (x || y) {
    if (x) out.println(A.next());
    if (y) out.println(B.next());
    x = A.hasNext();
    y = B.hasNext();
}
144
Check if file exists
Set boolean b to true if file at path fp exists on filesystem; false otherwise.

Beware that you should not do this and then in the next instruction assume the result is still valid, this is a race condition on any multitasking OS. 
boolean b = new File(fb).exists();
145
Print log line with datetime
Print message msg, prepended by current date and time.

Explain what behavior is idiomatic: to stdout or stderr, and what the date format is. 
Logger LOGGER = Logger.getLogger(MyClass.class.getName());

LOGGER.info(msg);
Alternative implementation:
out.printf("%tc:  `%s`%n", currentTimeMillis(), msg);
146
Convert string to floating point number
Extract floating point value f from its string representation s 
double f = Double.parseDouble(s);
Alternative implementation:
Float f = Float.valueOf(s);
Alternative implementation:
Double f = Double.valueOf(s);
147
Remove all non-ASCII characters
Create string t from string s, keeping only ASCII characters 
String t = s.replaceAll("[^\\x00-\\x7F]", "");
Alternative implementation:
String t = "";
char c;
int i, n = s.length();
for (i = 0; i < n; ++i)
    if ((c = s.charAt(i)) < 0x80)
        t = t + c;
Alternative implementation:
String t = s.chars()
    .filter(c -> c < 0x80)
    .mapToObj(x -> valueOf((char) x))
    .collect(joining());
148
Read list of integers from stdin
Read a list of integer numbers from the standard input, until EOF. 
List<Integer> list = new ArrayList<>();
Scanner in = new Scanner(System.in);
while(in.hasNext())
    list.add(in.nextInt());
Alternative implementation:
Scanner s = new Scanner(System.in);
int a[] = s.tokens()
    .mapToInt(Integer::parseInt)
    .toArray();
s.close();
149
Rescue the princess
As an exception, this content is not under license CC BY-SA 3.0 like the rest of this website.
150
Remove trailing slash
Remove the last character from the string p, if this character is a forward slash / 
if (p.endsWith("/")) {
    p = p.substring(0, p.length() - 1);
}
Alternative implementation:
p = p.replaceAll("/$", "");
Alternative implementation:
int n = p.length() - 1;
if (n != 0 && p.charAt(n) == '/')
    p = p.substring(0, n);
Alternative implementation:
p = p.replaceAll("(?<!^)/+$", "");
151
Remove string trailing path separator
Remove last character from string p, if this character is the file path separator of current platform.

Note that this also transforms unix root path "/" into the empty string! 
int n = p.length() - 1;
if (n != 0 && p.charAt(n) == separatorChar)
    p = p.substring(0, n);
152
Turn a character into a string
Create string s containing only the character c. 
String s = String.valueOf(c);
Alternative implementation:
String s = Character.toString(c);
Alternative implementation:
String s = c + "";
Alternative implementation:
String s = = new Character(c).toString();
Alternative implementation:
String s = "%s".formatted(c);
153
Concatenate string with integer
Create the string t as the concatenation of the string s and the integer i. 
String t = s + i;
Alternative implementation:
String t = "%s%s".formatted(s, i);
154
Halfway between two hex color codes
Find color c, the average between colors c1, c2.

c, c1, c2 are strings of hex color codes: 7 chars, beginning with a number sign # .
Assume linear computations, ignore gamma corrections. 
String r1 = c1.substring(1,3);
String g1 = c1.substring(3,5);
String b1 = c1.substring(5,7);

String r2 = c2.substring(1,3);
String g2 = c2.substring(3,5);
String b2 = c2.substring(5,7);

String r = String.format("%02X", (Integer.parseInt(r1, 16)+Integer.parseInt(r2, 16))/2 );
String g = String.format("%02X", (Integer.parseInt(g1, 16)+Integer.parseInt(g2, 16))/2 );
String b = String.format("%02X", (Integer.parseInt(b1, 16)+Integer.parseInt(b2, 16))/2 );

String c = "#" + r + g + b;
Alternative implementation:
StringBuilder sb = new StringBuilder("#");
for(int i=0;i<3;i++) {
  String sub1 = c1.substring(1+2*i,3+2*i);
  String sub2 = c2.substring(1+2*i,3+2*i);
  int v1 = Integer.parseInt(sub1, 16);
  int v2 = Integer.parseInt(sub2, 16);
  int v = (v1 + v2)/2;
  String sub = String.format("%02X", v);
  sb.append(sub);
}
String c = sb.toString();
Alternative implementation:
Color x = decode(c1.substring(1)),
      y = decode(c2.substring(1));
int r = (x.getRed()   + y.getRed())   / 2,
    g = (x.getGreen() + y.getGreen()) / 2,
    b = (x.getBlue()  + y.getBlue())  / 2;
String c = "#%02x%02x%02x".formatted(r, g, b);
155
Delete file
Delete from filesystem the file having path filepath. 
new File(filepath).delete();
Alternative implementation:
deleteIfExists(of(filepath));
156
Format integer with zero-padding
Assign to the string s the value of the integer i in 3 decimal digits. Pad with zeros if i < 100. Keep all digits if i ≥ 1000. 
String s = String.format("%03d", i);
Alternative implementation:
NumberFormat f = getIntegerInstance();
f.setMinimumIntegerDigits(3);
f.setGroupingUsed(false);
String s = f.format(i);
Alternative implementation:
String s = "%03d".formatted(i);
157
Declare constant string
Initialize a constant planet with string value "Earth". 
static final String planet = "Earth";
Alternative implementation:
final String planet = "Earth";
Alternative implementation:
final String planet = "Earth";
158
Random sublist
Create a new list y from randomly picking exactly k elements from list x.

It is assumed that x has at least k elements.
Each element must have same probability to be picked.
Each element from x must be picked at most once.
Explain if the original ordering is preserved or not. 
List<T> y = new ArrayList<>(x);
shuffle(y);
y = y.subList(0, k);
160
Detect if 32-bit or 64-bit architecture
Execute f32() if platform is 32-bit, or f64() if platform is 64-bit.
This can be either a compile-time condition (depending on target) or a runtime detection. 
switch(System.getProperty("sun.arch.data.model")) {
  case "64": 
    f64(); 
    break;
  case "32":
    f32();
    break;
}
161
Multiply all the elements of a list
Multiply all the elements of the list elements by a constant c 
elements = elements.stream().map(e -> e*c).collect(Collectors.toList());
Alternative implementation:
elements = elements.stream()
                   .map(x -> x * c)
                   .toList();
162
Execute procedures depending on options
execute bat if b is a program option and fox if f is a program option. 
var argsSet = Set.of(args);
if (argsSet.contains("b")) bat();
if (argsSet.contains("f")) fox();
Alternative implementation:
public static void main(String[] args) {
    for (String a : new LinkedHashSet<>(of(args)))
        switch (a) {
            case "b" -> bat();
            case "f" -> fox();
        }
}
163
Print list elements by group of 2
Print all the list elements, two by two, assuming list length is even. 
for(int i = 0; i < list.length; i += 2) {
  System.out.println(list[i] + ", " + list[i + 1]);
}
Alternative implementation:
range(0, list.size())
    .filter(i -> i % 2 == 0)
    .mapToObj(i -> list.get(i) + ", " + list.get(i + 1))
    .forEach(out::println);
165
Last element of list
Assign to the variable x the last element of the list items. 
int x = items[items.length - 1];
Alternative implementation:
T x = items.getLast();
166
Concatenate two lists
Create the list ab containing all the elements of the list a, followed by all the elements of the list b. 
var ab = new ArrayList<>(a);
ab.addAll(b);
Alternative implementation:
List<?> ab = new ArrayList<>(a) {{ addAll(b); }};
167
Trim prefix
Create the string t consisting of the string s with its prefix p removed (if s starts with p). 
String t = s.replaceFirst("^" + p, "");
Alternative implementation:
int i = s.indexOf(p), n = p.length();
t = i == 0 ? s.substring(n) : s;
Alternative implementation:
String t = s.replaceFirst('^' + quote(p), "");
168
Trim suffix
Create string t consisting of string s with its suffix w removed (if s ends with w). 
String t = s;
int index = s.lastIndexOf(w);
if (index + w.length() == s.length()) {
    t = s.substring(0, index);
}
Alternative implementation:
int i = s.lastIndexOf(w),
    m = s.length(), n = w.length();
t = i == m - n ? s.substring(0, i) : s;
Alternative implementation:
String t = s.replaceAll(quote(w) + '$', "");
169
String length
Assign to the integer n the number of characters of the string s.
Make sure that multibyte characters are properly handled.
n can be different from the number of bytes of s. 
int n = s.length();
170
Get map size
Set n to the number of elements stored in mymap.

This is not always equal to the map capacity. 
int n = mymap.size();
171
Add an element at the end of a list
Append the element x to the list s. 
s.add(x);
172
Insert an entry in a map
Insert value v for key k in map m. 
m.put(k, v);
173
Format a number with grouped thousands
Number will be formatted with a comma separator between every group of thousands. 
String.format("%,d", 1000000);
Alternative implementation:
NumberFormat f = new DecimalFormat("#,###");
String s = f.format(1_234);
Alternative implementation:
DecimalFormat f = (DecimalFormat) getNumberInstance();
f.setGroupingSize(3);
String s = f.format(1_000);
Alternative implementation:
out.printf("%,d", new BigInteger("1234"));
Alternative implementation:
out.printf("%,f", new BigDecimal("1234.5"));
Alternative implementation:
out.printf("%,f", 1_234.5);
174
Make HTTP POST request
Make a HTTP request with method POST to the URL u 
String s = HttpClient.newHttpClient().send(HttpRequest.newBuilder()
                        .uri(URI.create(u))
                        .POST(HttpRequest.BodyPublishers.ofString(content))
                        .build(), HttpResponse.BodyHandlers.ofString())
                .body();
175
Bytes to hex string
From the array a of n bytes, build the equivalent hex string s of 2n digits.
Each byte (256 possible values) is encoded as two hexadecimal characters (16 possible values per digit). 
String s = "";
for (byte b : a)
    s = s + "%02x".formatted(b);
Alternative implementation:
var s = HexFormat.of().formatHex(a);
Alternative implementation:
String s = iterate(0, x -> x + 1)
    .limit(n)
    .mapToObj(x -> a[x])
    .map("%02x"::formatted)
    .collect(joining());
176
Hex string to byte array
From hex string s of 2n digits, build the equivalent array a of n bytes.
Each pair of hexadecimal characters (16 possible values per digit) is decoded into one byte (256 possible values). 
byte[] a = new BigInteger(s, 16).toByteArray();
Alternative implementation:
public static byte[] hexToByteArray(String s) {
        int len = s.length();
        byte[] data = new byte[len / 2];
        for (int i = 0; i < len; i += 2) {
            data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
                    + Character.digit(s.charAt(i+1), 16));
        }
        return data;
    }
Alternative implementation:
int i, n = s.length();
byte a[] = new byte[n / 2];
for (i = 0; i < n; i = i + 2)
    a[i / 2] = (byte) fromHexDigits(s, i, i + 2);
177
Find files for a list of filename extensions
Construct a list L that contains all filenames that have the extension ".jpg" , ".jpeg" or ".png" in directory D and all its subdirectories. 
static List<String> L = new ArrayList<>();
record Find(File d, String ... t) {
    Find { get(d, t); }
    void get(File d, String ... t) {
        File a[] = d.listFiles();
        if (a == null) return;
        for (File f : a)
            if (f.isDirectory()) get(f, t);
            else asList(t).forEach(s -> {
                if (f.getName().endsWith(s))
                    L.add(f.getName());
            });
    }
}
178
Check if point is inside rectangle
Set boolean b to true if if the point with coordinates (x,y) is inside the rectangle with coordinates (x1,y1,x2,y2) , or to false otherwise.
Describe if the edges are considered to be inside the rectangle. 
boolean b = x <= x2 && x >= x1 && y <= y2 && y >= y1;
Alternative implementation:
double w = abs(x2 - x1), h = abs(y2 - y1);
Rectangle2D r = new Rectangle2D.Double(x1, y1, w, h);
boolean b = r.contains(x, y);
179
Get center of a rectangle
Return the center c of the rectangle with coördinates(x1,y1,x2,y2) 
double[] c = {(x1+x2)/2,(y1+y2)/2};
Alternative implementation:
double w = abs(x2 - x1), h = abs(y2 - y1), X, Y;
Rectangle2D r = new Rectangle2D.Double(x1, y1, w, h);
X = r.getCenterX();
Y = r.getCenterY();
Point2D c = new Point2D.Double(X, Y);
180
List files in directory
Create the list x containing the contents of the directory d.

x may contain files and subfolders.
No recursive subfolder listing. 
final File directory = new File(d);
final File[] x = directory.listFiles();
Alternative implementation:
String x[] = new File(d).list();
184
Tomorrow
Assign to variable t a string representing the day, month and year of the day after the current date. 
String t = LocalDate.now().plusDays(1).toString();
185
Execute function in 30 seconds
Schedule the execution of f(42) in 30 seconds. 
ScheduledExecutorService executorService = Executors.newSingleThreadScheduledExecutor();
executorService.schedule(() -> _f(42), delayInSeconds, TimeUnit.SECONDS);
Alternative implementation:
Thread t = new Thread(() -> {
    try {
        sleep(30_000);
        f(42);
    } catch (InterruptedException e) {
        throw new RuntimeException(e);
    }
});
t.start();
186
Exit program cleanly
Exit a program cleanly indicating no error to OS 
System.exit(0);
187
Disjoint Set
Disjoint Sets hold elements that are partitioned into a number of disjoint (non-overlapping) sets. 
public class DisjointSetElement {
  private DisjointSetElement representative = this;

  public DisjointSetElement getRepresentative() {
    if (representative != this) {
      representative = representative.getRepresentative();
    }
    return representative;
  }

  public void union(DisjointSetElement other) {
    other.getRepresentative().representative = getRepresentative();
  }
}
189
Filter and transform list
Produce a new list y containing the result of the function T applied to all elements e of the list x that match the predicate P. 
x.stream().filter(P).map(T).collect(Collectors.toList());
Alternative implementation:
List<T> y = x.stream()
             .filter(P)
             .map(T)
             .toList();
191
Check if any value in a list is larger than a limit
Given a one-dimensional array a, check if any value is larger than x, and execute the procedure f if that is the case 
for(int i = 0; i<a.length;i++) {
	if(a[i]>x) {
		f();
		break;
	}
}
Alternative implementation:
for (int i : a)
    if (i > x) {
        f();
        break;
    }
Alternative implementation:
if (of(a).anyMatch(i -> i > x)) f();
192
Declare a real variable with at least 20 digits
Declare a real variable a with at least 20 digits; if the type does not exist, issue an error at compile time. 
String pi = "3.14159265358979323846";
MathContext m = new MathContext(21, HALF_UP);
BigDecimal a = new BigDecimal(pi, m);
193
Transpose a two-dimensional matrix
Declare two two-dimensional arrays a and b of dimension n*m and m*n, respectively. Assign to b the transpose of a (i.e. the value with index interchange). 
<T> T[][] transpose(T[][] a) {
    int m, n, x = a[0].length, y = a.length;
    Class<?> t = a.getClass().getComponentType()
                             .getComponentType();
    T[][] b = (T[][]) newInstance(t, x, y);
    for (m = 0; m < y; ++m)
        for (n = 0; n < x; ++n)
            b[n][m] = a[m][n];
    return b;
}
195
Pass a two-dimensional array
Pass an array a of real numbers to the procedure (resp. function) foo. Output the size of the array, and the sum of all its elements when each element is multiplied with the array indices i and j (assuming they start from one). 
public static void foo(int[][] a){
       System.out.println("Array a has a size of " + a.length+" by " + a[0].length);
       int sum = 0;
       for (int i = 0; i<a.length; i++){
           for (int j = 0; j<a[0].length; j++){
               sum += ((i+1)*(j+1)*a[i][j]);
           }
       }
       System.out.println("The sum of all elements multiplied by their indices is: " + sum);
    }
Alternative implementation:
void foo(double a[][]) {
    int i, j, m = a.length, n;
    double z = 0, t;
    out.println(m);
    for (i = 0; i < m; ++i) {
        n = a[i].length;
        for (j = 0; j < n; ++j) {
            z = (z + ((t = a[i][j]) * (i + 1)));
            z = (z + (t * (j + 1)));
        }
    }
    out.println(z);
}
196
Pass a sub-array
Given an integer array a of size n, pass the first, third, fifth and seventh, ... up to the m th element to a routine foo which sets all these elements to 42. 
void foo(int a[], int m) {
    for (int i = 1; i <= m; a[i] = 42, i = i + 2);
}
Alternative implementation:
interface F { void set(int a[], int m); }
F f = (a, m) -> {
    for (int i = 1; i <= m; a[i] = 42, i = i + 2);
};
197
Get a list of lines from a file
Retrieve the contents of file at path into a list of strings lines, in which each element is a line of the file. 
List<String> lines = Files.readAllLines(new File(path).toPath());
Alternative implementation:
String text = Files.readString(new File(path).toPath());
List<String> lines = text.lines().collect(Collectors.toList());
Alternative implementation:
File f = new File(path);
List<String> lines = readAllLines(f.toPath());
Alternative implementation:
Scanner s = new Scanner(new File(path));
s.useDelimiter("\\R");
List<String> lines = s.tokens().toList();
s.close();
Alternative implementation:
FileReader R = new FileReader(path);
BufferedReader r = new BufferedReader(R);
List<String> lines = r.lines().toList();
r.close();
198
Abort program execution with error condition
Abort program execution with error condition x (where x is an integer value) 
System.exit(x);
200
Return hypotenuse
Compute the hypotenuse h of the triangle where the sides adjacent to the square angle have lengths x and y. 
return double h = Math.sqrt(Math.pow(x,2)+Math.pow(y,2));
Alternative implementation:
double h = hypot(x, y);
Alternative implementation:
double h = sqrt((x * x) + (y * y));
Alternative implementation:
BigDecimal X = new BigDecimal(x).pow(2),
           Y = new BigDecimal(y).pow(2),
           h = X.add(Y).sqrt(DECIMAL128);
201
Euclidean norm
Calculate n, the Euclidean norm of data (an array or list of floating point values). 
double n = 0d;
for(double value : data) {
	n += value * value;
}
n = Math.sqrt(n);
202
Sum of squares
Calculate the sum of squares s of data, an array of floating point values. 
double s = Arrays.stream(data).map(i -> i * i).sum();
Alternative implementation:
double s = of(data).map(x -> x * x).sum();
Alternative implementation:
BigDecimal s = of(data)
    .mapToObj(String::valueOf)
    .map(BigDecimal::new)
    .map(x -> x.multiply(x))
    .reduce(BigDecimal::add)
    .get();
203
Calculate mean and standard deviation
Calculate the mean m and the standard deviation s of the list of floating point values data. 
int i, n = data.length;
double u = data[0], g, t;
for (i = 1; i < n; ++i) u = u + data[i];
u = u / n;
for (g = i = 0; i < n; ++i)
    g = g + (t = data[i] - u) * t;
g = sqrt(g / n);
205
Get an environment variable
Read an environment variable with the name "FOO" and assign it to the string variable foo. If it does not exist or if the system does not support environment variables, assign a value of "none". 
String foo = System.getenv("foo");
if (foo == null) {
	foo = "none";
}
Alternative implementation:
String foo;
try {
    foo = getenv("FOO");
    if (foo == null) throw new Exception();
} catch (Exception e) {
    foo = "none";
}
206
Switch statement with strings
Execute different procedures foo, bar, baz and barfl if the string str contains the name of the respective procedure. Do it in a way natural to the language. 
switch (str){
	case "bar":
            bar();
            break;
        case "baz":
            baz();
            break;
        case "foo":
            foo();
            break;
        case "barfl":
            barfl();
            break;
        default:
            somethingElse();
    }
Alternative implementation:
switch (str) {
    case "foo" -> foo();
    case "bar" -> bar();
    case "baz" -> baz();
    case "barfl" -> barfl();
}
208
Formula with arrays
Given the arrays a,b,c,d of equal length and the scalar e, calculate a = e*(a+b*c+cos(d)).
Store the results in a. 
range(0, a.length)
    .forEach(i -> {
        a[i] = e * (a[0] + (b[1] * c[2]) + cos(d[3]));
    });
Alternative implementation:
int i, n = a.length;
for (i = 0; i < n; ++i)
    a[i] = e * (a[0] + (b[1] * c[2]) + cos(d[3]));
210
Compiler version and options
Assign, at runtime, the compiler version and the options the program was compiled with to variables version and options, respectively, and print them. For interpreted languages, substitute the version of the interpreter.

Example output:

GCC version 10.0.0 20190914 (experimental)
-mtune=generic -march=x86-64
 
public static void main(String[] args) {
    String version = getProperty("java.vm.version"),
           options = join(" ", args);
}
211
Create folder
Create the folder at path on the filesystem 
boolean ok = new File(path).mkdirs();
212
Check if folder exists
Set the boolean b to true if path exists on the filesystem and is a directory; false otherwise. 
File f = new File(path);
boolean b = f.exists() && f.isDirectory();
215
Pad string on the left
Prepend extra character c at the beginning of string s to make sure its length is at least m.
The length is the number of characters, not the number of bytes. 
while (s.length() < m) s = c + s;
216
Pad a string on both sides
Add the extra character c at the beginning and ending of string s to make sure its length is at least m.
After the padding the original content of s should be at the center of the result.
The length is the number of characters, not the number of bytes.

E.g. with s="abcd", m=10 and c="X" the result should be "XXXabcdXXX". 
char a[] = new char[m];
int n = s.length(), i = m / 2 - (n / 2), t = i - 1;
arraycopy(s.toCharArray(), 0, a, i, n);
while (--i != -1) a[i] = c;
while (--m != t + n) a[m] = c;
s = new String(a);
218
List intersection
Create the list c containing all unique elements that are contained in both lists a and b.
c should not contain any duplicates, even if a and b do.
The order of c doesn't matter. 
Set<T> A = copyOf(a), B = copyOf(b);
List<T> c = new ArrayList<>();
for (T t : A) if (B.contains(t)) c.add(t);
Alternative implementation:
a.retainAll(b);
List<T> c = List.copyOf(Set.copyOf(a));
Alternative implementation:
a.retainAll(b);
List<T> c = new ArrayList<>(copyOf(a));
Alternative implementation:
Set<T> A = copyOf(a), B = copyOf(b);
List<T> c = A.stream().filter(B::contains).toList();
219
Replace multiple spaces with single space
Create the string t from the value of string s with each sequence of spaces replaced by a single space.

Explain if only the space characters will be replaced, or the other whitespaces as well: tabs, newlines. 
String t = s.replaceAll(" +", " ");
Alternative implementation:
String t = s.replaceAll("\\s+", " ");
Alternative implementation:
String t = s.replaceAll(" {2,}", " ");
Alternative implementation:
String t = join(" ", s.split(" +", -1));
220
Create a tuple value
Create t consisting of 3 values having different types.

Explain if the elements of t are strongly typed or not. 
record Tuple(int a, String b, boolean c) {}
var t = new Tuple(1, "hello", true);
Alternative implementation:
class Tuple {
    String s;
    int i;
    boolean b;
}
Alternative implementation:
record Tuple<A, B, C>(A a, B b, C c) {}
Tuple<String, Integer, Boolean> t
    = new Tuple<>("abc", 123, true);
Alternative implementation:
record Tuple(Object ... a) {}
Tuple t = new Tuple("abc", 123, true);
221
Remove all non-digits characters
Create string t from string s, keeping only digit characters 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. 
StringBuilder tb = new StringBuilder();
for (int i=0; i<s.length(); i++){
    if(Character.isDigit(s.charAt(i)))
        tb.append(s.charAt(i));
}
String t = tb.toString();
Alternative implementation:
String t = s.replaceAll("\\D+", "");
Alternative implementation:
String t = "";
for (char c : s.toCharArray())
    if (isDigit(c)) t = t + c;
Alternative implementation:
String t = "";
char c;
for (int i = 0, n = s.length(); i < n; ++i)
    if (isDigit(c = s.charAt(i))) t = t + c;
Alternative implementation:
String t = s.chars()
    .filter(Character::isDigit)
    .mapToObj(x -> valueOf((char) x))
    .collect(joining());
222
Find the first index of an element in list
Set i to the first index in list items at which the element x can be found, or -1 if items does not contain x. 
int i = -1;
for(int j=0;j<items.length;j++){
	if(items[j].equals(x)){
		i = j;
		break;
	}
}
Alternative implementation:
int i = items.indexOf​(x);
223
for else loop
Loop through list items checking a condition. Do something else if no matches are found.

A typical use case is looping through a series of containers looking for one that matches a condition. If found, an item is inserted; otherwise, a new container is created.

These are mostly used as an inner nested loop, and in a location where refactoring inner logic into a separate function reduces clarity. 
boolean b = false;
for (int i = 0; i<items.length; i++) {
	if (items[i] == something) {
		doSomething();
		b = true;
	}
}
if (!b)
	doSomethingElse();
Alternative implementation:
Predicate<Item> condition;
items.stream()
	.filter(condition)
	.findFirst()
	.ifPresentOrElse(
		item -> doSomething(i),
		() -> doSomethingElse());
Alternative implementation:
 a: {
     b: {
            int i, n = items.size();
            for (i = 0; i < n; ++i)
                if (f(items.get(i)))
                    break b;
            out.println("not found");
            break a;
        }
        out.println("found");
    }
224
Add element to the beginning of the list
Insert the element x at the beginning of the list items. 
items.add(0, x);
225
Declare and use an optional argument
Declare an optional integer argument x to procedure f, printing out "Present" and its value if it is present, "Not present" otherwise 
private void f(Integer x) {
    if (x != null) {
        System.out.println("Present " + x);
    } else {
        System.out.println("Not present");
    }
}
Alternative implementation:
void f(int ... x) {
    if (x.length != 0) out.println("Present");
    else out.println("Not present");
}
Alternative implementation:
void f() { set(null); }
void f(int x) { set(x); }
private void set(Integer t) {
    if (t != null) out.println("Present");
    else out.println("Not present");
}
226
Delete last element from list
Remove the last element from the list items. 
items.remove(items.size() - 1);
Alternative implementation:
items.removeLast();
Alternative implementation:
items = copyOfRange(items, 0, items.length - 1);
Alternative implementation:
int n, t[] = new int[n = items.length - 1];
arraycopy(items, 0, t, 0, n);
items = t;
227
Copy a list
Create the new list y containing the same elements as the list x.

Subsequent modifications of y must not affect x (except for the contents referenced by the elements themselves if they contain pointers). 
List<T> y = new ArrayList<>(x);
228
Copy a file
Copy the file at path src to dst. 
Files.copy(Path.of(src), Path.of(dst));
229
Cancel an operation
Interrupt an ongoing processing p. 
p.interrupt();
230
Timeout
Cancel an ongoing processing p if it has not finished after 5s. 
Thread p = new Thread(() -> {});
p.start();
p.join(5_000);
p.interrupt();
232
Read a command line boolean flag
Print "verbose is true" if the flag -v was passed to the program command line, "verbose is false" otherwise. 
public static void main(String[] args) {
    boolean b = false;
    for (String s : args)
        if (s.equals("-v")) {
            b = true;
            break;
        }
    out.printf("verbose is %s", b);
}
233
Read a command line string flag
Print the value of the flag -country passed to the program command line, or the default value "Canada" if no such flag was passed. 
public static void main(String[] args) {
    String s;
    int i, n = args.length;
    for (i = 0; i < n; ++i)
        if (args[i].equals("-country"))
            break;
    s = i + 1 < n ? args[i + 1] : "Canada";
    out.println(s);
}
234
Encode bytes to base64
Assign to the string s the standard base64 encoding of the byte array data, as specified by RFC 4648. 
byte data[] = s.getBytes();
String s = getEncoder().encodeToString(data);
235
Decode base64
Assign to byte array data the bytes represented by the base64 string s, as specified by RFC 4648. 
byte[] data = Base64.getDecoder().decode(s);
236
Large quotient
Initialize a quotient q = a/b of arbitrary precision. a and b are large integers. 
BigDecimal A = new BigDecimal(a),
           B = new BigDecimal(b),
           q = A.divide(B);
237
Xor integers
Assign to c the result of (a xor b) 
int c = a ^ b;
238
Xor byte arrays
Write in a new byte array c the xor result of byte arrays a and b.

a and b have the same size. 
int i, n = a.length;
byte c[] = new byte[n];
for (i = 0; i < n; ++i)
    c[i] = (byte) (a[i] ^ b[i]);
Alternative implementation:
int n = a.length;
byte c[] = new byte[n];
iterate(0, i -> i + 1)
    .limit(n)
    .forEach(i -> {
        c[i] = (byte) (a[i] ^ b[i]);
    });
239
Find first regular expression match
Assign to string x the first word of string s consisting of exactly 3 digits, or the empty string if no such match exists.

A word containing more digits, or 3 digits as a substring fragment, must not match. 
Matcher m = compile("\\b\\d{3}\\b").matcher(s);
String x = m.find() ? m.group() : "";
241
Yield priority to other threads
Explicitly decrease the priority of the current process, so that other execution threads have a better chance to execute now. Then resume normal execution and call the function busywork. 
Thread.yield();
busywork();
242
Iterate over a set
Call a function f on each element e of a set x. 
for(T e: x) {
  f(e);
}
243
Print list
Print the contents of the list or array a on the standard output. 
System.out.println(a);
Alternative implementation:
System.out.println(Arrays.toString(a));
Alternative implementation:
out.println(deepToString(a));
244
Print a map
Print the contents of the map m to the standard output: keys and values. 
System.out.println(m);
Alternative implementation:
m.forEach((k, v) -> {
    out.println(k + " = " + v);
});
Alternative implementation:
m.entrySet().forEach(out::println);
245
Print value of custom type
Print the value of object x having custom type T, for log or debug. 
record X<T>(T t) {
    @Override
    public String toString() {
        String s = t.getClass().getName();
        s = "%s<%s>".formatted(getClass().getName(), s);
        return "%s@%x".formatted(s, identityHashCode(this));
    }
}
X<?> x = new X<>(123);
out.println(x);
Alternative implementation:
out.println(x);
246
Count distinct elements
Set c to the number of distinct elements in the list items. 
long c = items.stream().distinct().count();
247
Filter list in-place
Remove all the elements from list x that don't satisfy the predicate p, without allocating a new list.
Keep all the elements that do satisfy p.

For languages that don't have mutable lists, refer to idiom #57 instead. 
x.removeIf(p.negate());
249
Declare and assign multiple variables
Define variables a, b and c in a concise way.
Explain if they need to have the same type. 
Object a = 42, b = "Hello", c = 5.;
Alternative implementation:
int a, b, c = b = a = 123;
250
Pick a random value from a map
Choose a value x from map m.
m must not be empty. Ignore the keys. 
int i = new Random().nextInt(m.size());
V x = new ArrayList<>(m.values()).get(i);
251
Parse binary digits
Extract integer value i from its binary string representation s (in radix 2)
E.g. "1101" -> 13 
int i = new BigInteger(s, 2).intValue();
Alternative implementation:
Integer i = Integer.valueOf(s, 2);
252
Conditional assignment
Assign to the variable x the string value "a" if calling the function condition returns true, or the value "b" otherwise. 
x = condition() ? "a" : "b";
253
Print stack trace
Print the stack frames of the current execution thread of the program. 
for (var e : currentThread().getStackTrace())
    out.println(e);
254
Replace value in list
Replace all exact occurrences of "foo" with "bar" in the string list x 
x = x.stream()
     .map(s -> s.equals("foo") ? "bar" : s)
     .toList();
Alternative implementation:
x.replaceAll(e -> e.equals("foo") ? "bar" : e);
Alternative implementation:
ListIterator<String> i = x.listIterator();
while (i.hasNext())
    if (i.next().equals("foo")) i.set("bar");
255
Print a set
Print the values of the set x to the standard output.
The order of the elements is irrelevant and is not required to remain the same next time. 
System.out.println(x);
256
Count backwards
Print the numbers 5, 4, ..., 0 (included), one line per number. 
for(int i=5 ; i>=0 ; i--) {
    System.out.println(i);
}
Alternative implementation:
for (int i = 5; i != -1; out.println(i--));
Alternative implementation:
iterate(5, i -> i - 1)
    .limit(6)
    .forEach(out::println);
257
Traverse list backwards
Print each index i and value x from the list items, from the last down to the first. 
for(int i = items.length-1;i>=0;i--){
	System.out.println("Index:"+ i +", Value:" + items[i]); 
}
Alternative implementation:
reverse(items);
int i = items.size() - 1;
for (T x : items)
    out.printf("%s %s%n", i--, x);
Alternative implementation:
T x;
int i = items.size() - 1;
while (i != -1) {
    x = items.get(i);
    out.printf("%s %s%n", i--, x);
}
Alternative implementation:
int n = items.size();
iterate(n - 1, i -> i - 1)
    .limit(n)
    .forEach(i -> {
        out.println(i + " " + items.get(i));
    });
258
Convert list of strings to list of integers
Convert the string values from list a into a list of integers b. 
b = a.stream()
     .map(Integer::parseInt)
     .toList();
Alternative implementation:
for (String s : a) b.add(parseInt(s));
259
Split on several separators
Build the list parts consisting of substrings of the input string s, separated by any of the characters ',' (comma), '-' (dash), '_' (underscore). 
List<?> parts = list(new StringTokenizer(s, ",-_"));
Alternative implementation:
String parts[] = s.split("[%s]".formatted(quote(",-_")), -1);
Alternative implementation:
Scanner t = new Scanner(s);
t.useDelimiter("[%s]".formatted(quote(",-_")));
String list[] = t.tokens().toArray(String[]::new);
t.close();
260
Create an empty list of strings
Declare a new list items of string elements, containing zero elements 
List<String> list = new ArrayList<>();
261
Format time hours-minutes-seconds
Assign to the string x the value of fields (hours, minutes, seconds) of the date d, in format HH:MM:SS. 
long d = currentTimeMillis();
String x = "%tH:%<tM:%<tS".formatted(d);
Alternative implementation:
Date d = getInstance().getTime();
String x = "%tT".formatted(d);
262
Count trailing zero bits
Assign to t the number of trailing 0 bits in the binary representation of the integer n.

E.g. for n=112, n is 1110000 in base 2 ⇒ t=4 
int t;
if (n % 2 != 0) t = 0;
else {
    String b = toBinaryString(n);
    t = b.replaceAll(".+?(?<!0)(?=0+)", "").length();
}
Alternative implementation:
int t = numberOfTrailingZeros(n);
Alternative implementation:
String b = toBinaryString(n);
int i = b.lastIndexOf('1'), t;
t = i == -1 ? 0 : b.substring(++i).length();
263
Integer logarithm in base 2
Write two functions log2d and log2u, which calculate the binary logarithm of their argument n rounded down and up, respectively. n is assumed to be positive. Print the result of these functions for numbers from 1 to 12. 
interface F {
    double log2 = log(2);
    double f(double n);
}
F log2d = x -> floor(log(x) / F.log2),
  log2u = x -> ceil(log(x) / F.log2);

for (int i = 1; i <= 12; ++i) {
    out.printf("log2d(%s) = %s%n", i, log2d.f(i));
    out.printf("log2u(%s) = %s%n", i, log2u.f(i));
}
264
Automated passing of array bounds
Pass a two-dimensional integer array a to a procedure foo and print the size of the array in each dimension. Do not pass the bounds manually. Call the procedure with a two-dimensional array. 
int i = 0, m = a.length;
while (i < m) out.println(a[i++].length);
265
Even parity bit
Calculate the parity p of the integer variable i : 0 if it contains an even number of bits set, 1 if it contains an odd number of bits set. 
int p = Integer.bitCount(i) % 2;
266
Repeated string
Assign to the string s the value of the string v repeated n times, and write it out.

E.g. v="abc", n=5 ⇒ s="abcabcabcabcabc" 
for (int i=0;i<n;i++){
          s = s+v;
      }
System.out.println(s);
Alternative implementation:
String s = v.repeat(n);

System.out.println(s);
Alternative implementation:
String s = generate(() -> v)
    .limit(n)
    .collect(joining());
267
Pass string to argument that can be of any type
Declare an argument x to a procedure foo that can be of any type. If the type of the argument is a string, print it, otherwise print "Nothing."

Test by passing "Hello, world!" and 42 to the procedure. 
    public static void main(String[] args) {
        foo("Hello, world!");
        foo(42);
    }

    private static void foo(Object x) {
        if (x instanceof String) {
            System.out.println(x);
        } else {
            System.out.println("Nothing.");
        }
    }
Alternative implementation:
<T> void foo(T x) {
    if (x instanceof String) out.println(x);
    else out.println("Nothing.");
}
Alternative implementation:
interface F<T> { void set(T x); }
F<Object> foo = x -> {
    if (x instanceof String) out.println(x);
    else out.println("Nothing.");
};
foo.set("Hello, world!");
foo.set(42);
269
Enum to String
Given the enumerated type t with 3 possible values: bike, car, horse.
Set the enum value e to one of the allowed values of t.
Set the string s to hold the string representation of e (so, not the ordinal value).
Print s. 
var e = T.HORSE;
var s = e.name();
System.out.println(s);
Alternative implementation:
T e = Car;
String s = e.name();
out.println(s);
270
Test for quiet or signaling NaN
Given a floating point number r1 classify it as follows:
If it is a signaling NaN, print "This is a signaling NaN."
If it is a quiet NaN, print "This s a quiet NaN."
If it is not a NaN, print "This is a number."
 
if (isNaN(r1)) out.println("This s a quiet NaN.");
else out.println("This is a number.");
Alternative implementation:
if (r1 != r1) out.println("This s a quiet NaN.");
else out.println("This is a number.");
271
Test for type extension
If a variable x passed to procedure tst is of type foo, print "Same type." If it is of a type that extends foo, print "Extends type." If it is neither, print "Not related." 
<T> void tst(T x) {
    Class<?> c = x.getClass();
    if (c.getSuperclass() == Foo.class)
        out.println("Extends type.");
    else if (c == Foo.class)
        out.println("Same type.");
    else
        out.println("Not related.");
}
272
Play FizzBuzz
Fizz buzz is a children's counting game, and a trivial programming task used to affirm that a programmer knows the basics of a language: loops, conditions and I/O.

The typical fizz buzz game is to count from 1 to 100, saying each number in turn. When the number is divisible by 3, instead say "Fizz". When the number is divisible by 5, instead say "Buzz". When the number is divisible by both 3 and 5, say "FizzBuzz" 
for (int i = 1; i <= 100; i++){
	if (i % 15 == 0)
	  System.out.println ("FizzBuzz");
	else if (i % 3 == 0)
	  System.out.println ("Fizz");
	else if (i % 5 == 0)
	  System.out.println ("Buzz");
	else
	  System.out.println (i);
      }
273
Check if folder is empty
Set the boolean b to true if the directory at filepath p is empty (i.e. doesn't contain any other files and directories) 
boolean b = p.list().length == 0;
274
Remove all white space characters
Create the string t from the string s, removing all the spaces, newlines, tabulations, etc. 
String t = s.replaceAll("\\s+", "");
Alternative implementation:
String t = "";
for (char c : s.toCharArray())
    if (!isWhitespace(c)) t = t + c;
Alternative implementation:
String t = "";
for (char c : s.toCharArray())
    switch (c) {
        case ' ', '\t', '\n', '\r' -> {}
        default -> t = t + c;
    }
Alternative implementation:
String t = s.chars()
    .filter(c -> !isWhitespace(c))
    .mapToObj(c -> valueOf((char) c))
    .collect(joining());
275
Binary digits to byte array
From the string s consisting of 8n binary digit characters ('0' or '1'), build the equivalent array a of n bytes.
Each chunk of 8 binary digits (2 possible values per digit) is decoded into one byte (256 possible values). 
int i, m = s.length(), n = (m + 1) / 8, t;
byte a[] = new byte[n];
for (i = 0; i < m; i = i + 8) {
    t = parseInt(s.substring(i, i + 8), 2);
    a[i / 8] = (byte) t;
}
276
Insert an element in a set
Insert an element e into the set x. 
x.add(e)
277
Remove an element from a set
Remove the element e from the set x.

Explains what happens if e was already absent from x. 
x.remove(e);
278
Read one line from the standard input
Read one line into the string line.

Explain what happens if EOF is reached. 
String line = null;
Scanner in = new Scanner(System.in);
if(in.hasNextLine())
    line = in.nextLine();
Alternative implementation:
Scanner s = new Scanner(System.in);
String line = s.nextLine();
s.close();
279
Read list of strings from the standard input
Read all the lines (until EOF) into the list of strings lines. 
List<String> lines = new ArrayList<>();
Scanner in = new Scanner(System.in);
while(in.hasNextLine())
    lines.add(in.nextLine());
280
Filter map
Remove all the elements from the map m that don't satisfy the predicate p.
Keep all the elements that do satisfy p.

Explain if the filtering happens in-place, i.e. if m is reused or if a new map is created. 
Iterator<?> i = m.entrySet().iterator();
while (i.hasNext()) 
    if (!p.test(i.next())) i.remove();
Alternative implementation:
m.entrySet().removeIf(e -> !p.test(e));
281
Use a Point as a map key
You have a Point with integer coordinates x and y. Create a map m with key type Point (or equivalent) and value type string. Insert "Hello" at position (42, 5). 
Map<Point, String> m = new HashMap<>();
m.put(new Point(42, 5), "Hello");
283
Split with a custom string separator
Build the list parts consisting of substrings of input string s, separated by the string sep. 
List<?> parts = list(new StringTokenizer(s, sep));
Alternative implementation:
String parts[] = s.split(quote(sep), -1);
284
Create a zeroed list of integers
Create a new list a (or array, or slice) of size n, where all elements are integers initialized with the value 0. 
int[] a = new int[n];
Alternative implementation:
int[] a = IntStream.generate(() -> 0)
                .limit(n)
                .toArray();
Alternative implementation:
List<Integer> a = new ArrayList<Integer>(n);
for (int i =0;i<n;i++){
    a.add(0);
}
Alternative implementation:
int a[] = new int[n];
Alternative implementation:
Integer a[] = new Integer[n];
fill(a, 0);
Alternative implementation:
List<Integer> a = new ArrayList<>(n);
range(0, n).forEach(i -> a.add(0));
Alternative implementation:
int a[] = iterate(0, i -> 0).limit(n).toArray();
Alternative implementation:
int a[] = range(0, n).map(i -> 0).toArray();
286
Iterate over characters of a string
Print a line "Char i is c" for each character c of the string s, where i is the character index of c in s (not the byte index).

Make sure that multi-byte characters are properly handled, and count for a single character. 
AtomicInteger i = new AtomicInteger(0);
s.chars().forEach(v -> {
  char c = (char)v;
  System.out.println("Char " + i.getAndIncrement() + " is " + c); 
});
Alternative implementation:
for(int i=0; i<s.length(); i++) {
  char c = s.charAt(i);
  System.out.println("Char " + i + " is " + c); 
}
Alternative implementation:
range(0, s.length())
    .forEach(i -> {
        out.printf("Char %s is %s%n", i, s.charAt(i));
    });
Alternative implementation:
record E(int c) {
    static int z;
    E { out.printf("Char %s is %c%n", z++, c); }
}
s.chars().forEach(E::new);
288
Check if set contains a value
Set the boolean b to true if the set x contains the element e, false otherwise. 
boolean b = x.contains(e);
289
Concatenate two strings
Create the string s by concatenating the strings a and b. 
String s = a.concat(b);
Alternative implementation:
String s = a + b;
Alternative implementation:
String s = "%s%s".formatted(a, b);
Alternative implementation:
StringBuilder t = new StringBuilder(a);
Formatter f = new Formatter(t);
f.format("%s", b).flush();
String s = t.toString();
290
Sort sublist
Sort the part of the list items from index i (included) to index j (excluded), in place, using the comparator c.

Elements before i and after j must remain unchanged. 
items.subList(i, j).sort(Comparable::compareTo);
291
Remove sublist
Delete all the elements from index i (included) to index j (excluded) from the list items. 
items.removeAll(items.subList(i, j));
293
Create a stack
Create a new stack s, push an element x, then pop the element into the variable y. 
Stack<T> s = new Stack<>();
s.push(x);
T y = s.pop();
294
Print a comma-separated list of integers
Given an array a containing the three values 1, 12, 42, print out
"1, 12, 42" with a comma and a space after each integer except the last one.
 
out.print(a[0]);
int i = 1, n = a.length;
while (i < n) out.print(", " + a[i++]);
Alternative implementation:
String s = of(a)
    .mapToObj(String::valueOf)
    .collect(joining(", "));
out.println(s);
Alternative implementation:
String s = of(a)
    .mapToObj(String::valueOf)
    .reduce((x, y) -> x + ", " + y)
    .get();
out.println(s);
295
String to Enum
Given the enumerated type T, create a function TryStrToEnum that takes a string s as input and converts it into an enum value of type T.

Explain whether the conversion is case sensitive or not.
Explain what happens if the conversion fails.
 
T x = T.valueOf(s);
296
Replace last occurrence of substring
Assign to x2 the value of string x with the last occurrence of y replaced by z.
If y is not contained in x, then x2 has the same value as x. 
String x2 = x;
int i = x.lastIndexOf(y);
if (i != -1) {
    String t = x2.substring(0, i);
    x2 = t + z + x2.substring(i + y.length());
}
Alternative implementation:
StringBuilder x2 = new StringBuilder(x);
int i = x2.lastIndexOf(y);
x2.replace(i, i + y.length(), z);
Alternative implementation:
String p = quote(y), x2;
x2 = x.replaceAll(p + "(?!.*" + p + ')', z);
297
Sort a list of strings, case-insensitively
Sort the string list data in a case-insensitive manner.

The sorting must not destroy the original casing of the strings. 
data.sort(String::compareToIgnoreCase);
298
Copy a map
Create the map y by cloning the map x.

y is a shallow copy, not a deep copy. 
Map<K, V> y = x;
299
Comment out a single line
Write a line of comments.

This line will not be compiled or executed. 
// This is a comment
302
String interpolation
Given the integer x = 8, assign to the string s the value "Our sun has 8 planets", where the number 8 was evaluated from x. 
String s = String.format("Our sun has %s planets", x);
Alternative implementation:
String s = "Our sun has %s planets".formatted(x);
Alternative implementation:
String f = "Our sun has {0,number} planets",
       s = format(f, x);
Alternative implementation:
Formatter f = new Formatter();
f.format("Our sun has %s planets", x);
f.flush();
String s = f.toString();
303
Array with non-default lower bound
Declare an array a of integers with six elements, where the first index is 42 and consecutive elements have the indices 43, 44, 45, 46, 47. 
int a[] = new int[42 + 6];
for (int i = 42, n = 42 + 6; i < n; ++i) {

}
304
Encode string into UTF-8 bytes
Create the array of bytes data by encoding the string s in UTF-8. 
byte data[] = s.getBytes(UTF_8);
305
Calculate exponentiation of real numbers
Compute and print a^b, and a^n, where a and b are floating point numbers and n is an integer.
 
out.println(pow(a, b));
out.println(pow(a, n));
306
Ensure list capacity
Preallocate memory in the list x for a minimum total capacity of 200 elements.

This is not possible in all languages. It is only meant as a performance optimization, should not change the length of x, and should not have any effect on correctness. 
List<T> x = new ArrayList<>(200);
Alternative implementation:
ArrayList<?> x = new ArrayList<>();
x.ensureCapacity(200);
308
Integer to string in base b
Create the string representation s of the integer value n in base b.

18 in base 3 -> "200"
26 in base 5 -> "101"
121 in base 12 -> "a1"

 
String s = Integer.toString(n, b);
309
Clone a 2D array
Create the new 2-dimensional array y containing a copy of the elements of the 2-dimensional array x.

x and y must not share memory. Subsequent modifications of y must not affect x. 
Class<?> c = x.getClass(),
         a = c.getComponentType(),
         b = a.getComponentType();
int i, m = x.length, n;
T y[][] = (T[][]) newInstance(a, m), Y[], t[];
for (i = 0; i < m; ++i) {
    n = (t = x[i]).length;
    Y = (T[]) newInstance(b, n);
    arraycopy(t, 0, Y, 0, n);
    y[i] = Y;
}
310
Fill array with random bytes
Fill the byte array a with randomly generated bytes. 
new Random().nextBytes(a);
312
List equality
Set b to true if the lists p and q have the same size and the same elements, false otherwise. 
b = p.equals(q);
313
Map equality
Set b to true if the maps m and n have the same key/value entries, false otherwise. 
boolean b = m.equals(n);
314
Fill array with value
Set all the elements in the array x to the same value v 
fill(x, v);
316
Count occurrences in a list
Determine the number c of elements in the list x that satisfy the predicate p. 
int c = frequency(x, p);
317
Random string
Create a string s of n characters having uniform random values out of the 62 alphanumeric values A-Z, a-z, 0-9 
String s = "";
Integer z[], i = 0;
char a[] = new char[62], c;
for (c = 'a'; c <= 'z'; ++c) a[i++] = c;
for (c = 'A'; c <= 'Z'; ++c) a[i++] = c;
for (c = '0'; c <= '9'; ++c) a[i++] = c;
z = new Integer[62];
for (i = 0; i < 62; ++i) z[i] = i;
shuffle(asList(z));
for (i = 0; i < n; ++i) s = s + a[z[i]];
319
Generator function
Write a function g that behaves like an iterator.
Explain if it can be used directly in a for loop. 
Random r = new Random();
Supplier<Integer> s = r::nextInt;
Stream<Integer> g = generate(s);
Alternative implementation:
Stream<?> g = iterate(0, i -> i + 1).limit(100);
320
Test if string is empty
Set b to true if the string s is empty, false otherwise 
boolean b = s.equals("");
Alternative implementation:
boolean b = s.intern() == "";
321
Access character in string, by index
Assign to c the value of the i-th character of the string s.

Make sure to properly handle multi-byte characters. i is the character index, which may not be equal to the byte index. 
char c = s.charAt(i);
325
Create a queue
Create a new queue q, then enqueue two elements x and y, then dequeue an element into the variable z. 
Deque<T> q = new ArrayDeque<>();
q.offer(x);
q.offer(y);
T z = q.poll();
Alternative implementation:
Deque<T> q = new ArrayDeque<>();
q.addLast(x);
q.addLast(y);
T z = q.removeFirst();
326
Milliseconds since Unix epoch
Assign to t the number of milliseconds elapsed since 00:00:00 UTC on 1 January 1970. 
long t = currentTimeMillis();
Alternative implementation:
long t = new Date().getTime();
Alternative implementation:
long t = Instant.now().toEpochMilli();
327
Convert string to lower case
Assign to t the value of the string s, with all letters mapped to their lower case. 
t = s.toLowerCase();
328
Convert string to upper case
Assign to t the value of the string s, with all letters mapped to their upper case. 
String t = s.toUpperCase();
329
Read value in a map
Assign to v the value stored in the map m for the key k.

Explain what happens if there is no entry for k in m. 
v = m.get(k);
Alternative implementation:
m.getOrDefault(k, null);
331
Clear map
Remove all entries from the map m.

Explain if other references to the same map now see an empty map as well. 
m.clear();
332
List of the keys of a map
Create the list k containing all the keys of the map m 
List<K> k = new ArrayList<>(m.keySet());
Alternative implementation:
List<K> k = copyOf(m.keySet());
334
Combine 2 maps
Create the new map c containing all of the (key, value) entries of the two maps a and b.

Explain what happens for keys existing in both a and b. 
Map<K, V> c = new HashMap<>(a);
c.putAll(b);
335
List to map
Create the map m containing all the elements e of the list a, using as key the field e.id. 
Map<Integer, E> m = new HashMap<>();
a.forEach(e -> m.put(e.id, e));
Alternative implementation:
Map<Integer, E> m = a.stream()
    .collect(toMap(e -> e.id, e -> e));
Alternative implementation:
Map<Integer, E> m = new HashMap<>();
for (E e : a) m.put(e.id, e);
336
Exponent
Compute x = b ⁿ

b raised to the power of n is equal to the product of n terms b × b × ... × b 
double x = pow(b, n);
337
Convert string to integer, explicit base
Extract the integer value i from its string representation s, in radix b 
Integer i = Integer.valueOf(s, b);
338
Create a bidirectional map
Create a new bidirectional map bm and insert the pair (42, "forty-two") 
BiMap<Integer, String> bm = HashBiMap.create();
bm.put(42, "forty-two");
339
Clear a byte array
Set all the elements of the byte array a to zero 
Arrays.fill(a, (byte)0);
340
Last character of string
Assign to c the value of the last character of the string s.

Explain the type of c, and what happens if s is empty.

Make sure to properly handle multi-bytes characters. 
int n = s.length(), i = n - 1;
if (isSurrogate(s.charAt(i))) --i;
String c = s.substring(i);
341
Find substring last position
Set i to the position of the last occurrence of the string y inside the string x, if exists.

Specify if i should be regarded as a character index or as a byte index.

Explain the behavior when y is not contained in x. 
int i = x.lastIndexOf(y);
342
Leap year?
Determine if the current year is a leap year. 
int y = parseInt("%tY".formatted(currentTimeMillis()));
boolean b = y % 4 == 0 && (y % 100 != 0 || y % 400 == 0);
Alternative implementation:
boolean b = Year.now().isLeap();
344
Extract filename extension
Assign to ext the fragment of the string f after the last dot character, or the empty string if f does not contain a dot.

E.g. "photo.jpg" -> "jpg"

ext must not contain the dot character. 
int i = f.lastIndexOf('.');
String ext = i != -1 ? f.substring(++i) : "";
345
Convert string to big integer
Create the integer value i initialized from its string representation s (in radix 10)

Use an integer type that can hold huge values. Explain what happens if s cannot be parsed. 
BigInteger i = new BigInteger(s, 10);
348
Convert a decimal value into a fraction
Parse a number, a, into a mathematical fraction, f.

For example, 0.5 is `1/2`, and 3.125 is `3 1/8`.

https://en.wikipedia.org/wiki/Fraction 
String s = valueOf(a), f;
s = s.substring(s.indexOf('.') + 1);
int n = parseInt(s);
if (n == 0) f = valueOf((int) a);
else {
    record GCF() {
        static int of(int a, int b) {
            return b == 0 ? a : of(b, a % b);
        }
    }
    int d = (int) pow(10, s.length()),
        gcf = GCF.of(n, d), i = (int) a;
    if (gcf != 0) {
        n = n / gcf;
        d = d / gcf;
    }
    f = n + "/" + d;
    if (i != 0) f = i + " " + f;
}
349
Convert a fraction into a decimal value
Parse a value, f, into a decimal number, a.

For example, `1/2` is 0.5, and `3 1/8` is 3.125.

https://en.wikipedia.org/wiki/Fraction 
int z = 0, n, d, i;
double a;
if ((i = f.indexOf(' ')) != -1) {
    z = parseInt(f.substring(0, i));
    f = f.substring(++i).trim();
}
i = f.indexOf('/');
n = parseInt(f.substring(0, i));
d = parseInt(f.substring(++i));
a = z + ((double) n / d);
350
Create an enumerated collection
Generate a collection of "enumerated" values.

For example, Python's `enumerate` built-in function.

https://docs.python.org/3/library/functions.html#enumerate 
record E<T>(int i, T value) {}
<T> List<E<T>> enumerate(T ... a) {
    Iterator<T> g = stream(a).iterator();
    return iterate(0, x -> g.hasNext(), i -> i + 1)
        .mapToObj(i -> new E<>(i, g.next()))
        .toList();
}
351
Create a zipped collection
Generate a "zipped" list z of pairs of elements from the lists a, b having the same length n.

The result z will contain n pairs. 
Object[] zip(Object[] ... c) {
    Object[] t, T;
    int a, b, i, m = c.length,
        z = stream(c).mapToInt(x -> x.length)
                     .max().getAsInt();
    t = new Object[z];
    for (b = i = 0; b < z; ++b) {
        T = new Object[m];
        for (a = 0; a < m; ++a)
            if (b < c[a].length)
                T[a] = c[a][b];
        t[i++] = T;
    }
    return t;
}