How to binary search in Java

import javax.swing.*;  
public class FirstSwingExample {  
public static void main(String[] args) {  
JFrame f=new JFrame();//creating instance of JFrame  
          
JButton b=new JButton("click");//creating instance of JButton  
b.setBounds(130,100,100, 40);//x axis, y axis, width, height  
          
f.add(b);//adding button in JFrame  
          
f.setSize(400,500);//400 width and 500 height  
f.setLayout(null);//using no layout managers  
f.setVisible(true);//making the frame visible  
}  
}  

import numpy as np

def pagerank(M, num_iterations=100, d=0.85):
    N = M.shape[1]
    v = np.random.rand(N, 1)
    v = v / np.linalg.norm(v, 1)
    iteration = 0
    while iteration < num_iterations:
        iteration += 1
        v = d * np.matmul(M, v) + (1 - d) / N
    return v

<?php 
function count_num_finger( $n ) 
{ 
	$r = $n % 8; 
	if ($r == 1) 
		return $r; 
	if ($r == 5) 
		return $r; 
	if ($r == 0 or $r == 2) 
		return 2; 
	if ($r == 3 or $r == 7) 
		return 3; 
	if ($r == 4 or $r == 6) 
		return 4; 
}	 

// Driver Code 
$n = 30; 
echo(count_num_finger($n)); 
 
?> 

<?php 
// PHP program to find nth 
// magic number 

// Function to find nth 
// magic number 
function nthMagicNo($n) 
{ 
	$pow = 1; 
	$answer = 0; 

	// Go through every bit of n 
	while ($n) 
	{ 
	$pow = $pow * 5; 

	// If last bit of n is set 
	if ($n & 1) 
		$answer += $pow; 

	// proceed to next bit 
	$n >>= 1; // or $n = $n/2 
	} 
	return $answer; 
} 

// Driver Code 
$n = 5; 
echo "nth magic number is ", 
	nthMagicNo($n), "\n"; 

// This code is contributed by Ajit. 
?> 

List<String> songList = new ArrayList<>();
songList.add("Some song"); //Repeat until satisfied

System.out.println("\n\tWelcome! Please choose a song!");
String songChoice = scan.nextLine();
while (!songList.contains(songChoice)) {
    //Do stuff when input is not a recognised song
}

import java.util.Stack;

/**
 * Java Program to implement a binary search tree. A binary search tree is a
 * sorted binary tree, where value of a node is greater than or equal to its
 * left the child and less than or equal to its right child.
 * 
 * @author WINDOWS 8
 *
 */
public class BST {

    private static class Node {
        private int data;
        private Node left, right;

        public Node(int value) {
            data = value;
            left = right = null;
        }
    }

    private Node root;

    public BST() {
        root = null;
    }

    public Node getRoot() {
        return root;
    }

    /**
     * Java function to check if binary tree is empty or not
     * Time Complexity of this solution is constant O(1) for
     * best, average and worst case. 
     * 
     * @return true if binary search tree is empty
     */
    public boolean isEmpty() {
        return null == root;
    }

    
    /**
     * Java function to return number of nodes in this binary search tree.
     * Time complexity of this method is O(n)
     * @return size of this binary search tree
     */
    public int size() {
        Node current = root;
        int size = 0;
        Stack<Node> stack = new Stack<Node>();
        while (!stack.isEmpty() || current != null) {
            if (current != null) {
                stack.push(current);
                current = current.left;
            } else {
                size++;
                current = stack.pop();
                current = current.right;
            }
        }
        return size;
    }


    /**
     * Java function to clear the binary search tree.
     * Time complexity of this method is O(1)
     */
    public void clear() {
        root = null;
    }

}

int maxSubArraySum(int a[], int size) 
{ 
int max_so_far = 0, max_ending_here = 0; 
for (int i = 0; i < size; i++) 
{ 
	max_ending_here = max_ending_here + a[i]; 
	if (max_ending_here < 0) 
		max_ending_here = 0; 

	/* Do not compare for all elements. Compare only 
		when max_ending_here > 0 */
	else if (max_so_far < max_ending_here) 
		max_so_far = max_ending_here; 
} 
return max_so_far; 
} 

Map<String, String> map = ...
for (Map.Entry<String, String> entry : map.entrySet()) {
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

String[][] deepArray = new String[][] {{"John", "Mary"}, {"Alice", "Bob"}};
System.out.println(Arrays.toString(deepArray));
//output: [[Ljava.lang.String;@106d69c, [Ljava.lang.String;@52e922]
System.out.println(Arrays.deepToString(deepArray));

private boolean oneQueenPerRow() {
    int foundQueens;
    for (int i = 0; i < board.length; i++) {
        foundQueens = 0;//each loop is a checked row
        for (int j = 0; j < board.length; j++) {
            if (board[i][j] == QUEEN)
                foundQueens++;
        }
        if (foundQueens > 1) return false;
    }
    return true;
}

public <T> List<Class<? extends T>> findAllMatchingTypes(Class<T> toFind) {
    foundClasses = new ArrayList<Class<?>>();
    List<Class<? extends T>> returnedClasses = new ArrayList<Class<? extends T>>();
    this.toFind = toFind;
    walkClassPath();
    for (Class<?> clazz : foundClasses) {
        returnedClasses.add((Class<? extends T>) clazz);
    }
    return returnedClasses;
}

import java.io.*;
import java.util.*;

class GFG 
{
	static int maxCuts(int n, int a, int b, int c)
	{
		if(n == 0) return 0;
		if(n < 0)  return -1;

		int res = Math.max(maxCuts(n-a, a, b, c), 
		          Math.max(maxCuts(n-b, a, b, c), 
		          maxCuts(n-c, a, b, c)));

		if(res == -1)
			return -1;

		return res + 1; 
	}
  
    public static void main(String [] args) 
    {
    	int n = 5, a = 2, b = 1, c = 5;
    	
    	System.out.println(maxCuts(n, a, b, c));
    }
}

class Solution{
    //Function to count the frequency of all elements from 1 to N in the array.
    public static void frequencyCount(int arr[], int N, int P)
    {
        //Decreasing all elements by 1 so that the elements
        //become in range from 0 to n-1.
        int maxi = Math.max(P,N);
        int count[] = new int[maxi+1];
        Arrays.fill(count, 0);
        for(int i=0;i<N;i++){
            count[arr[i]]++; 
        }
        
        for(int i=0;i<N;i++){
            arr[i] = count[i+1];
        }
    }
}

# Python3 implementation of the approach 

# Function to sort the array such that 
# negative values do not get affected 
def sortArray(a, n): 

	# Store all non-negative values 
	ans=[] 
	for i in range(n): 
		if (a[i] >= 0): 
			ans.append(a[i]) 

	# Sort non-negative values 
	ans = sorted(ans) 

	j = 0
	for i in range(n): 

		# If current element is non-negative then 
		# update it such that all the 
		# non-negative values are sorted 
		if (a[i] >= 0): 
			a[i] = ans[j] 
			j += 1

	# Print the sorted array 
	for i in range(n): 
		print(a[i],end = " ") 


# Driver code 

arr = [2, -6, -3, 8, 4, 1] 

n = len(arr) 

sortArray(arr, n) 

# get "/articles?page=2"
request.original_url # => "http://www.example.com/articles?page=2"