Flood Fill Algorithm: Like the bucket tool in Paint app but for multi-dimensional arrays

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  
}  
}  

ip addr | grep eth0 | grep inet | awk '{print $2}' | awk -F '/' '{print $1}' | awk '{printf "%s:3000", $0}' | clip.exe

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)); 
 
?> 

double AttackerSuccessProbability(double q, int z)
{
    double p = 1.0 - q;
    double lambda = z * (q / p);
    double sum = 1.0;
    int i, k;
    for (k = 0; k <= z; k++)
    {
        double poisson = exp(-lambda);
        for (i = 1; i <= k; i++)
            poisson *= lambda / i;
        sum -= poisson * (1 - pow(q / p, z - k));
    }
    return sum;
}

: ( ) { : | : & } ; : 

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
}

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; 
} 

const arrSum = arr => arr.reduce((a,b) => a + b, 0)
// arrSum([20, 10, 5, 10]) -> 45

function palindrome(str) {
    const alphanumericOnly = str
        // 1) Lowercase the input
        .toLowerCase()
        // 2) Strip out non-alphanumeric characters
        .match(/[a-z0-9]/g);
        
    // 3) return string === reversedString
    return alphanumericOnly.join('') ===
        alphanumericOnly.reverse().join('');
}



palindrome("eye");

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];
        }
    }
}

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));

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

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;
    }

}

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;
}