Trapping Rain Water

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  
}  
}  

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
}

add_filter( 'woocommerce_product_get_stock_quantity' ,'custom_get_stock_quantity', 10, 2 );
add_filter( 'woocommerce_product_variation_get_stock_quantity' ,'custom_get_stock_quantity', 10, 2 );
function custom_get_stock_quantity( $value, $product ) {
    $value = 15; // <== Just for testing
    return $value;
}

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];
        }
    }
}

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));
    }
}

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;
}

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.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;

class Pair {
    int x, y;

    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

class FloodFill
{
    // Below arrays details all 8 possible movements
    private static final int[] row = { -1, -1, -1, 0, 0, 1, 1, 1 };
    private static final int[] col = { -1, 0, 1, -1, 1, -1, 0, 1 };

    // check if it is possible to go to pixel (x, y) from
    // current pixel. The function returns false if the pixel
    // has different color or it is not a valid pixel
    public static boolean isSafe(char[][] M, int m, int n,
                                int x, int y, char target)
    {
        return x >= 0 && x < m && y >= 0 && y < n
                && M[x][y] == target;
    }

    // Flood fill using BFS
    public static void floodfill(char[][] M, int x, int y, char replacement)
    {
        int m = M.length;
        int n = M[0].length;

        // create a queue and enqueue starting pixel
        Queue<Pair> q = new ArrayDeque<>();
        q.add(new Pair(x, y));

        // get target color
        char target = M[x][y];

        // run till queue is not empty
        while (!q.isEmpty())
        {
            // pop front node from queue and process it
            Pair node = q.poll();

            // (x, y) represents current pixel
            x = node.x;
            y = node.y;

            // replace current pixel color with that of replacement
            M[x][y] = replacement;

            // process all 8 adjacent pixels of current pixel and
            // enqueue each valid pixel
            for (int k = 0; k < row.length; k++)
            {
                // if adjacent pixel at position (x + row[k], y + col[k]) is
                // a valid pixel and have same color as that of current pixel
                if (isSafe(M, m, n, x + row[k], y + col[k], target))
                {
                    // enqueue adjacent pixel
                    q.add(new Pair(x + row[k], y + col[k]));
                }
            }
        }
    }

    public static void main(String[] args)
    {
        // matrix showing portion of the screen having different colors
        char[][] M = {
            "YYYGGGGGGG".toCharArray(),
            "YYYYYYGXXX".toCharArray(),
            "GGGGGGGXXX".toCharArray(),
            "WWWWWGGGGX".toCharArray(),
            "WRRRRRGXXX".toCharArray(),
            "WWWRRGGXXX".toCharArray(),
            "WBWRRRRRRX".toCharArray(),
            "WBBBBRRXXX".toCharArray(),
            "WBBXBBBBXX".toCharArray(),
            "WBBXXXXXXX".toCharArray()
        };

        // start node
        int x = 3, y = 9;   // target color = "X"

        // replacement color
        char replacement = 'C';

        // replace target color with replacement color
        floodfill(M, x, y, replacement);

        // print the colors after replacement
        for (int i = 0; i < M.length; i++) {
            System.out.println(Arrays.toString(M[i]));
        }
    }
}

# 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)