The snippets are under the CC-BY-SA license.

Creative Commons Attribution-ShareAlike 3.0

Logo

Programming-Idioms.org

  • The snippets are under the CC-BY-SA license.
  • Please consider keeping a bookmark
  • (instead of printing)
Java
1
Print a literal string on standard output
System.out.println("Hello World");
Alternative implementation:
System.out.printf("%s", "Hello, World!");
2
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:
for(int count = 1; count < 11; count++) {
	System.out.printf("%s%n", "Hello");
}
3
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:
private static void methodName() {
	System.out.println("Hello, World!");
}
4
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:
private int square(int value) {
	return value * value;
}
5
Declare a container type for two floating-point numbers x and y
class Point{
  double x;
  double y;
}
Alternative implementation:
private class Point
{
	private final float x;
	private final float y;
	
	public Point(float x, float y)
	{
		this.x = x;
		this.y = y;
	}
	
	public float getX()
	{
		return this.x;
	}
	
	public float getY()
	{
		return this.y;
	}
}
6
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 item : items) {
	System.out.println(item);
}
7
Print each index i with its value x from an array-like collection items
for (int i = 0; i < items.length; i++) {
    T x = items[i];
    System.out.printf("Item %d = %s%n", i, x);
}
Alternative implementation:
for (int i = 0; i < items.size(); i++) {
    T x = items.get(i);
    System.out.printf("Item %d = %s%n", i, x);
}
Alternative implementation:
int i = 0;
for(String x : items) {
	System.out.printf("%d %s%n", i, x);
	i++;
}
8
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);
}};
9
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
Generate a random permutation of the elements of list x
Collections.shuffle(x);
11
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 != null && x.size() > 0) {
	Random random = new Random();
	T item = x.get(random.nextInt(x.size()));
}
12
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)
13
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:
for(Map.Entry<K, V> item : mymap.entrySet()) {
	K k = item.getKey();
	V x = item.getValue();
	System.out.printf("%s, %s%n", k, x);
}
14
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 random = new Random();
float value = random.nextFloat(a, b);
Alternative implementation:
Random random = new Random();
float value = (random.nextFloat() * (b - a)) + a;
15
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);
}
16
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
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 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);
20
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;
}
21
Swap the values of the variables a and b
T tmp = a;
a = b;
b = tmp;
22
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);
23
Given a real number x, create its string representation s with 2 decimal digits following the dot.
String s = String.format("%.2f", x);
24
Declare a new string s and initialize it with the literal value "ネコ" (which means "cat" in japanese)
String s = "ネコ";
25
Share the string value "Alan" with an existing running process which will then display "Hello, Alan"
https://gist.github.com/f7c174c53efaba4d8575
26
Declare and initialize a matrix x having m rows and n columns, containing real numbers.
double[][] x = new double[m][n];
27
Declare and initialize a 3D array x, having dimensions boundaries m, n, p, and containing real numbers.
double[][][] x = new double[m][n][p];
28
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))
29
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);
30
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
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
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
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);
}
34
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);
35
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);
36
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
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
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
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);
40
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
Create string t containing the same characters as string s, in reverse order.
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 = "";
char[] characters = s.toCharArray();
for(int index = s.length() - 1; index > -1; index--) {
	t += String.valueOf(characters[index]);
}
42
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
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;
		}
44
Insert the element x at position i in the list s. Further elements must be shifted to the right.
s.add(i, x);
45
Sleep for 5 seconds in current thread, before proceeding with the next instructions.
Thread.sleep(5000);
Alternative implementation:
TimeUnit.SECONDS.sleep(5);
46
Create string t consisting of the 5 first characters of string s.
Make sure that multibyte characters are properly handled.
String t = s.substring(0,5);
47
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);
48
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 lineSeparator = System.lineSeparator();
String s = "line1" + lineSeparator + "line2" + lineSeparator;
49
Build list chunks consisting in substrings of the string s, separated by one or more space characters.
String[] chunks = s.split("\\s+");
Alternative implementation:
List<String> chunks = Arrays.asList(s.split("\\s+"));
50
Write a loop that has no end clause.
for(;;){
	// Do something
}
Alternative implementation:
while(true) {
	// Do something	
}
51
Determine whether the map m contains an entry for the key k
m.containsKey(k)
52
Determine whether the map m contains an entry with the value v, for some key.
m.containsValue(v)
53
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 = "";
for(int index = 0; index < x.size(); index++) {
	if(index != 0) {
		y += ", ";
	}
	y += x.get(index);
}
54
Calculate the sum s of the integer list or array x.
int s = 0;
for (int i : x) {
  s += i;
}
55
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 = String.format("%d", i);
56
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.awaitTermination(10L, TimeUnit.MINUTES);
System.out.println("Finished");
57
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.
var y = x.stream().filter(p).collect(Collectors.toList());
58
Create 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);
59
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
Assign to x the string value of the first command line parameter, after the program name.
String x = args[0];
61
Assign to the variable d the current date/time value, in the most standard type.
Instant d = Instant.now();
Alternative implementation:
long unixEpoch = System.currentTimeMillis();
String d = String.format("%1$tY-%1$tm-%1$td %1$tI:%1$tM:%1$tS", unixEpoch);
Alternative implementation:
long d = System.currentTimeMillis();
62
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
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
Assign to x the value 3^247
BigInteger x = new BigInteger("3", 10).pow(247);
65
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 = String.format("%.1f%%", x * 100f);
66
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
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 an object x to store n bits (n being potentially large).
BitSet x = new BitSet(n);
69
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
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
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));
    }
}
74
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 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
Create the string s of integer x written in base 2.

E.g. 13 -> "1101"
String s = Integer.toBinaryString(x);
77
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
Execute a block once, then execute it again as long as boolean condition c is true.
do {
	someThing();
	someOtherThing();
} while(c);
79
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));
80
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
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);
82
Find how many times string s contains substring t.
Specify if overlapping occurrences are counted.
int count = StringUtils.countMatches(s, t);
Alternative implementation:
int count = 0;
Pattern pattern = Pattern.compile(String.format("(%s)", t));
Matcher matcher = pattern.matcher(s);
while(matcher.find()) {
	count++;
}
Alternative implementation:
int count = 0;
Pattern pattern = Pattern.compile(String.format("(%s)", t), Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(s);
while(matcher.find()) {
	count++;
}
83
Declare regular expression r matching strings "http", "htttp", "httttp", etc.
Pattern r = Pattern.compile("htt+p");
84
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 = 0;
for(char character : Integer.toBinaryString(i).toCharArray()) {
	if(character == '1') {
		c++;
	}
}
85
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;
}
86
Write boolean function multiplyWillOverflow 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.
static boolean multiplyWillOverflow(int x, int y) {
	return Integer.MAX_VALUE/x < y;
}
87
Exit immediately.
If some extra cleanup work is executed by the program runtime (not by the OS itself), describe it.
System.exit(0);
88
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];
89
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);
90
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;
    }
}
93
Implement the procedure control which receives one parameter f, and runs f.
static void control(Runnable f) {
    f.run();
}
94
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:
System.out.print(x.getClass().getName());
if(x.getClass().getModifiers() == Modifier.STATIC) {
	System.out.println(" is static");
} else {
	System.out.println(" is dynamic");
}
95
Assign to variable x the length (number of bytes) of the local file at path.
long x = new File(path).length();
96
Set boolean b to true if string s starts with prefix prefix, false otherwise.
boolean b = s.startsWith(prefix);
97
Set boolean b to true if string s ends with string suffix, false otherwise.
boolean b = s.endsWith(suffix);
98
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 = System.currentTimeMillis();
String d = String.format("%1$tY-%1$tm-%1$td %1$tI:%1$tM:%1$tS", ts);
99
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)
100
Sort elements of array-like collection items, using a comparator c.
Arrays.sort(items, c);
Alternative implementation:
Collections.sort(items, c);
101
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();
105
Assign to the string s the name of the currently executing program (but not its full path).
String s = System.getProperty("sun.java.command");
106
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
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.)
public class Example
{
	public static int x = 1;
	
	public static void printFieldX()
	{
		for(Field field : Example.class.getFields()) {
			if(field.getName().equals("x")) {
				System.out.print(x);
			}
		}
	}
}
109
Set n to the number of bytes of a variable t (of type T).
String className = t.getClass().getSimpleName();
int n = 0;
if (className.equals("Byte")) {
	n = 1;
} else if (className.equals("Short")) {
	n = 2;
} else if (className.equals("Integer")) {
	n = 4;
} else if (className.equals("Long")) {
	n = 8;
} else if (className.equals("Float")) {
	n = 4;
} else if (className.equals("Double")) {
	n = 8;
} else if (className.equals("Boolean")) {
	n = 0;
} else if (className.equals("Character")) {
	n = 2;
}
110
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.equals("") || s.matches("\\s+");
111
From current process, run program x with command-line parameters "a", "b".
Runtime.getRuntime().exec(new String[]{"x", "a", "b"});
112
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);
115
Set boolean b to true if date d1 is strictly before date d2 ; false otherwise.
boolean before = (d1.compareTo(d2) == -1);
116
Remove all occurrences of string w from string s1, and store the result in s2.
String s2 = s1.replace(w, "");
117
Set n to the number of elements of the list x.
int n = x.size();
Alternative implementation:
int n = x.length;
118
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);
119
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);
  }
}
120
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();
121
Listen UDP traffic on port p and read 1024 bytes into 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
Create an enumerated type Suit with 4 possible values SPADES, HEARTS, DIAMONDS, CLUBS.
enum Suit{
  SPADES,
  HEARTS,
  DIAMONDS,
  CLUBS;
}
123
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
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;
}
125
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");
126
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:
private static class Tuple<S, B>
{
	private S s;
	private B b;
	
	public Tuple(S s, B b)
	{
		this.setS(s);
		this.setB(b);
	}
	
	public void setS(S s)
	{
		this.s = s;
	}
	
	public void setB(B b)
	{
		this.b = b;
	}
	
	public S getS()
	{
		return s;
	}
	
	public B getB()
	{
		return b;
	}
}

private static Tuple foo()
{
	return new Tuple<String, Boolean>("value", true);
}
128
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
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();
}
132
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
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());
134
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);
135
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;
	}
}
136
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));
137
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]+");
138
Create a new temporary file on the filesystem.
File tempFile = File.createTempFile("prefix-", "-suffix");
tempFile.deleteOnExit();
139
Create a new temporary folder on filesystem, for writing.
File folder = new File("/path/");
folder.deleteOnExit();
140
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 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 (T item : items1) {
	System.out.println(item);
}
for (T item : items2) {
	System.out.println(item);
}
Alternative implementation:
Stream.concat(
	items1.stream(), 
	items2.stream()
).forEach(System.out::println);
142
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 = String.format("%02x", x);
143
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 index = 0;
String value1 = null;
String value2 = null;
while(index < items1.size() && index < items2.size()) {
	if(index < items1.size()) {
		value1 = items1.get(index);
	}
	if(index < items2.size()) {
		value2 = items2.get(index);
	}
	System.out.printf("%s, %s%n", value1, value2);
	index++;
}
144
Set boolean b to true if file at path fp exists on filesystem; false otherwise.

Beware that you should never 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 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:
long unixEpoch = System.currentTimeMillis();
String dateAndTime = String.format("%1$tY-%1$tm-%1$td %1$tI:%1$tM:%1$tS", unixEpoch);
System.out.printf("%s: %s%n", dateAndTime, msg);
System.err.printf("%s: %s%n", dateAndTime, msg);
146
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
Create string t from string s, keeping only ASCII characters
String t = s.replaceAll("[^\\x00-\\x7F]", "");
148
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());
149
As an exception, this content is not under license CC BY-SA 3.0 like the rest of this website.
 
150
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("/$", "");
152
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 = String.format("%s", c);
153
Create the string t as the concatenation of the string s and the integer i.
String t = s + i;
Alternative implementation:
String t = String.format("%s%d", s, i);
154
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:
String c1 = "#152a3e";
String c2 = "#005867";
int cx = Integer.parseInt(c1.substring(1), 16);
int cy = Integer.parseInt(c2.substring(1), 16);
int r1 = (cx & 0xff_00_00) >> 16;
int g1 = (cx & 0x00_ff_00) >> 8;
int b1 = (cx & 0x00_00_ff);
int r2 = (cy & 0xff_00_00) >> 16;
int g2 = (cy & 0x00_ff_00) >> 8;
int b2 = (cy & 0x00_00_ff);
int r = (r1 + r2) / 2;
int g = (g1 + g2) / 2;
int b = (b1 + b2) / 2;
String c = String.format("#%02x%02x%02x", r, g, b);
155
Delete from filesystem the file having path filepath.
new File(filepath).delete();
156
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 i1000.
String s = String.format("%03d", i);
157
Initialize a constant planet with string value "Earth".
static final String planet = "Earth";
Alternative implementation:
final String planet = "Earth";
Alternative implementation:
private final String planet = "Earth";
158
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.
public static <T> List<T> randomSublist(List<T> x, int k) {
	List<T> y = new ArrayList<>(x);
	Collections.shuffle(y);
	return y.subList(0, k);
}
160
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 the list elements by a constant c
elements = elements.stream().map(e -> e*c).collect(Collectors.toList());
162
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();
163
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:
for(int index1 = 0, index2 = 1; index2 < list.size(); index1++, index2 = index1 + 1) {
	System.out.printf("%s, %s%n", list.get(index1), list.get(index2));
}
165
Assign to the variable x the last element of the list items.
int x = items[items.length - 1];
166
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);
167
Create string t consisting of string s with its prefix p removed (if s starts with p).
String t = s.replaceFirst("^" + p, "");
168
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);
}
169
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
Set n to the number of elements stored in mymap.

This is not always equal to the map capacity.
int n = mymap.size();
171
Append the element x to the list s.
s.add(x);
172
Insert value v for key k in map m.
m.put(k, v);
173
Number will be formatted with a comma separator between every group of thousands.
String.format("%,d", 1000000);
Alternative implementation:
DecimalFormat decimalFormat = new DecimalFormat("#,###");
String formattedString = decimalFormat.format(1000000);
174
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
From 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 value : a) {
	s += String.format("%02x", value);
}
176
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();
177
Construct a list L that contains all filenames that have the extension ".jpg" , ".jpeg" or ".png" in directory D and all its subdirectories.
public static ArrayList<String> findFiles(String D)
{
	File fileDirectory = new File(D);
	ArrayList<String> L = new ArrayList<>();
	
	String fileName = null;
	for(File file : fileDirectory.listFiles()) {
		fileName = file.getName();
		if(fileName.endsWith(".jpg") || fileName.endsWith(".jpeg") || fileName.endsWith(".png")) {
			L.add(fileName);
		}
	}
	
	return L;
}
178
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;
179
Return the center c of the rectangle with coördinates(x1,y1,x2,y2)
double[] c = {(x1+x2)/2,(y1+y2)/2};
180
Create list x containing the contents of 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:
ArrayList<File> x = (ArrayList<File>) Arrays.asList(new File(d).listFiles());
184
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
Schedule the execution of f(42) in 30 seconds.
ScheduledExecutorService executorService = Executors.newSingleThreadScheduledExecutor();
executorService.schedule(() -> _f(42), delayInSeconds, TimeUnit.SECONDS);
186
Exit a program cleanly indicating no error to OS
System.exit(0);
187
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
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());
191
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 value : a) {
	if(value > x) {
		f();
		break;
	}
}
195
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);
    }
197
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());
198
Abort program execution with error condition x (where x is an integer value)
System.exit(x);
200
Returns 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));
201
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
Calculate the sum of squares s of data, an array of floating point values.
double s = Arrays.stream(data).map(i -> i * i).sum();
205
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";
}
206
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();
    }
211
Create the folder at path on the filesystem
boolean ok = new File(path).mkdirs();
215
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.
if(s.length() < m)
    s = String.valueOf(c).repeat(m - s.length()) + s;
216
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".
if(s.length() < m) {
    String x = String.valueOf(c).repeat(m - s.length());
    s = x.substring(0, x.length() / 2) + s + x.substring(x.length() / 2);
}
219
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+", " ");
220
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);
221
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("[^0-9]+, "");
222
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
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());
224
Insert the element x at the beginning of the list items.
items.add(0, x);
225
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");
    }
}
226
Remove the last element from list items.
items.remove(items.size() - 1);
227
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).
ArrayList<String> y = new ArrayList<>(x);
228
Copy the file at path src to dst.
Files.copy(Path.of(src), Path.of(dst));
235
Assign to byte array data the bytes represented by the base64 string s, as specified by RFC 4648.
byte[] data = Base64.getDecoder().decode(s);
237
Assign to c the result of (a xor b)
int c = a ^ b;
238
Write in a new byte array c the xor result of byte arrays a and b.

a and b have the same size.
byte[] c = new byte[a.length];
for (int i = 0; i < a.length; i++) {
  c[i] = (byte) (a[i] ^ b[i]);
}
239
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.
Pattern pattern = Pattern.compile("((?<!\\d)\\d{3}(?!\\d))");
Matcher matcher = pattern.matcher(s);
String x = "";
matcher.find();
x = matcher.group(1);
241
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 function busywork.
Thread.yield();
busywork();
242
Call a function f on each element e of a set x.
for(T e: x) {
  f(e);
}
243
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));
244
Print the contents of the map m to the standard output: keys and values.
System.out.println(m);
246
Set c to the number of distinct elements in the list items.
long c = items.stream().distinct().count();
247
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());
251
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
Assign to the variable x the value "a" if calling the function condition returns true, or the value "b" otherwise.
x = condition() ? "a" : "b";
253
Print the stack frames of the current execution thread of the program.
for(StackTraceElement stackTraceElement : Thread.currentThread().getStackTrace()) {
	System.out.println(stackTraceElement.toString());
}
254
Replace all exact occurrences of "foo" with "bar" in the string list x
x = x.stream()
     .map(s -> s.equals("foo") ? "bar" : s)
     .collect(Collectors.toList());
255
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
Print the numbers 5, 4, ..., 0 (included), one line per number.
for(int i=5 ; i>=0 ; i--) {
    System.out.println(i);
}
257
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:
Collections.reverse(items);
int i = 0;
for(T x : items) {
	System.out.printf("%d %s%n", i, x);
	i++;
}
258
Convert the string values from list a into a list of integers b.
List<Integer> b = items.stream()
                .map(Integer::parseInt)
                .collect(Collectors.toList());
Alternative implementation:
ArrayList<Integer> b = new ArrayList<>();
for(String value : a) {
	b.add(Integer.parseInt(value));
}
260
Declare a new list items of string elements, containing zero elements
List<String> list = new ArrayList<>();
262
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
Pattern pattern = Pattern.compile("(0+)$");
Matcher matcher = pattern.matcher(Integer.toBinaryString(n));
int t = 0;
if(matcher.find()) {
	t = matcher.group(1).length();
}
Alternative implementation:
Integer t = Integer.numberOfTrailingZeros(n);
265
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
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);
267
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.");
        }
    }
269
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);
272
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);
      }
274
Create the string t from the string s, removing all the spaces, newlines, tabulations, etc.
String t = s.replaceAll("\\s+", "");
275
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).
short[] a = new short[s.length() / 8];
for(int index1 = 0, index2 = 0; index1 < a.length; index1++, index2 += 8) {
	a[index1] = Short.valueOf(s.substring(index2, index2 + 8), 2);
}
276
Insert an element e into the set x.
x.add(e)
277
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 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:
InputStreamReader inputStreamReader = new InputStreamReader(System.in);
BufferedReader bufferedReader = new BufferedReader(inputStreamReader);

String line = null;
try {
	line = bufferedReader.readLine();
} catch (IOException exception) {
	exception.printStackTrace();
} finally {
	try {
		bufferedReader.close();
	} catch(IOException exception) {
		exception.printStackTrace();
	}
}
279
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());
281
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");
284
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);
}
286
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:
int i = 0;
for(char c : s.toCharArray()) {
	System.out.printf("Char %d is %c%n", i, c);
	i++;
}
288
Set the boolean b to true if the set x contains the element e, false otherwise.
boolean b = x.contains(e);
289
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 = String.format("%s%s", a, b);
290
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, j and after, must remain unchanged.
Comparator<T> c = new Comparator<T>()
{
	@Override
	public int compare(T o1, T o2)
	{
		if(o1 > o2) {
			return 1;
		} else if(o1 == o2) {
			return 0;
		} else {
			return -1;
		}
	}
};
Collections.sort(items.subList(i, j), c);
291
Delete all the elements from index i (included) to index j (excluded) from the list items.
for(int index = i; index < j; index++) {
	items.remove(index);
}
293
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
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.
int[] a = new int[] { 1, 12, 42 };
for(int index = 0; index < a.length; index++) {
	if(index != 0) {
		System.out.print(", ");
	}
	System.out.print(a[index]);
}
295
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);
297
Sort the string list data in a case-insensitive manner.

The sorting must not destroy the original casing of the strings.
data.sort(new Comparator<String>()
{
	@Override
	public int compare(String o1, String o2)
	{
		return o1.compareToIgnoreCase(o2);
	}
	
});
298
Create the map y by cloning the map x.

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

This line will not be compiled or executed.
// This is a comment
302
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);
303
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[6];
for(int value = 42, index = 0; value < 48; value++, index++) {
	a[index] = value;
}
308
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
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.
int[][] y = new int[x.length][];
for(int index1 = 0; index1 < x.length; index1++) {
	y[index1] = new int[x[index1].length];
	for(int index2 = 0; index2 < x[index1].length; index2++) {
		y[index1][index2] = x[index1][index2];
	}
}
310
Fill the byte array a with randomly generated bytes.
new Random().nextBytes(a);
312
Set b to true if the lists p and q have the same size and the same elements, false otherwise.
boolean b = p.size() == q.size();

if(b) {
	for(T item : p) {
		if(q.contains(item) == false) {
			b = false;
			break;
		}
	}
	
	if(b) {
		for(T item : q) {
			if(p.contains(item) == false) {
				b = false;
				break;
			}
		}
	}
}
316
Determine the number c of elements in the list x that satisfy the predicate p.
int c = Collections.frequency(x, p);
317
Create a string s of n characters having uniform random values out of the 62 alphanumeric values A-Z, a-z, 0-9
public static String generateValues(int n)
{
	String alphabet1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
	String alphabet2 = "abcdefghijklmnopqrstuvwxyz";
	String digits = "0123456789";
	String values = alphabet1 + alphabet2 + digits;
		
	Random random = new Random(1);
	String s = "";
	for(int count = 0; count < n; count++) {
		s += String.valueOf(values.charAt(random.nextInt(values.length())));
	}
	
	return s;
}
320
Set b to true if the string s is empty, false otherwise
boolean b = s.equals("");
321
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);