Kadane’s Algorithm

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)); 
 
?> 

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;




void decToBinary(int n)
{
    // array to store binary number 
    int binaryNum[3][3];
    
    //converting to binary 
    for (int i = 0; i < 3; i++) 
    {
        for (int j = 0; j < 3; j++)
        {
            binaryNum[i][j] = n % 2;
            n = n / 2;

        }
     }       

    // printing binary> array in reverse order 
    for (int i = 3-1; i >= 0; i--){
        for (int j = 3 - 1; j >= 0; j--)
        {
            if (binaryNum[i][j] == 0)
                cout << "H" << " ";
            else
                cout << "T" << " ";
        }
        cout << endl;
    }
    
 }

int main()
{
    int n;
    cout << "Enter a decimal number between 1 and 512 ";
    cin >> n;

    decToBinary(n);
    return 0;
}

#include<iostream>
using namespace std;
bool issafe(int** arr,int x,int y,int n){
    for(int row=0;row<=x;row++){
        if(arr[x][y]==1){
            return false;
        }
    }
    int row=x;
    int col=y;
    while(row>=0 && col>=0){
        if(arr[row][col]==1){
            return false;
        }
        row--;
        col--;
    }
     row=x;
     col=y;
    while(row>=0 && col<n){
        if(arr[row][col]==1){
            return false;
        }
        row--;
        col++;
    }
    return true ;
}
bool nqueen(int** arr,int x,int n){
    if(x>=n){
        return true;
    }
    for(int col=0;col<=n;col++){
        if(issafe(arr,x,col,n)){
            arr[x][col]==1;
            if(nqueen(arr,x+1,n)){
                return true;
            }
            arr[x][col]=0;  //backtracking
            
        }
    }
    return false;
}
int main(){
    int n;
    cin>>n;
    int** arr=new int*[n];
    for(int i=0;i<n;i++){
        arr[i]=new int[n];
        for(int j=0;j<n;j++){
            arr[i][j]=0;

        }
    }
    if(nqueen(arr,0,n)){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cout<<arr[i][j]<<"";
            }cout<<endl;
        }
    }


    return 0;
}

<?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. 
?> 

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];
        }
    }
}

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;


bool primeNumber(int n);
bool Emirp(int n);
int reversal(int n);
int recursive(int a, int b);

int main()
{
	
	int count = 0;
	int number = 13;
	while (count <= 100)
	{
		if (Emirp(number))
		{
			count++;
			if (count % 10 == 0)
				cout << setw(7) << number << endl;
			else
				cout << setw(7) << number;
		}
		number++;
	}
		

}
bool primeNumber(int n) {
	
		for (int i = 2; i <= n / 2; i++) {
			
			if (n % i == 0) {

				return false;
			}
		}
		return true;
}

bool Emirp(int n) {
	

	return primeNumber(n) && primeNumber(reversal(n));
	
}

int reversal(int n) {
	
		if (n < 10) {
			
				return n;
			
		}
		return recursive(n % 10, n / 10);
	
}

int recursive(int a, int b) {

	if (b < 1) {

		return a;

	}
	return recursive(a * 10 + b % 10, b / 10);
	
}



	
   

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;
    }

}

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)