Subset array sum by generating all the subsets

Given an array of size N and a sum, the task is to check whether some array elements can be added to sum to N . 
Note: At least one element should be included to form the sum.(i.e. sum cant be zero) 
Examples:

Input: array = -1, 2, 4, 121, N = 5
Output: YES
The array elements 2, 4, -1 can be added to sum to N

Input: array = 1, 3, 7, 121, N = 5
Output:NO 

Approach: The idea is to generate all subsets using Generate all subsequences of array and correspondingly check if any subsequence has the sum equal to the given sum.
Below is the implementation of the above approach: 

CPP

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Find way to sum to N using array elements atmost once
void find(int arr[], int length, int s)
{
    // loop for all 2^n combinations
    for (int i = 1; i < (pow(2, length)); i++) {

        // sum of a combination
        int sum = 0;

        for (int j = 0; j < length; j++)

            // if the bit is 1 then add the element
            if (((i >> j) & 1))
                sum += arr[j];

        // if the sum is equal to given sum print yes
        if (sum == s) {
            cout << "YES" << endl;
            return;
        }
    }

    // else print no
    cout << "NO" << endl;
}

// driver code
int main()
{
    int sum = 5;
    int array[] = { -1, 2, 4, 121 };
    int length = sizeof(array) / sizeof(int);

    // find whether it is possible to sum to n
    find(array, length, sum);

    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

        // Find way to sum to N using array elements atmost once
        static void find(int [] arr, int length, int s)
        {
            // loop for all 2^n combinations
            for (int i = 1; i <= (Math.pow(2, length)); i++) {

                // sum of a combination
                int sum = 0;

                for (int j = 0; j < length; j++)

                    // if the bit is 1 then add the element
                    if (((i >> j) & 1) % 2 == 1)
                        sum += arr[j];

                // if the sum is equal to given sum print yes
                if (sum == s) {
                    System.out.println("YES");
                    return;
                }
            }

            // else print no
            System.out.println("NO");
        }

        // driver code
        public static void main(String[] args)
        {
            int sum = 5;
            int []array = { -1, 2, 4, 121 };
            int length = array.length;

            // find whether it is possible to sum to n
            find(array, length, sum);

        }

}

// This code is contributed by ihritik

Python3

# Python3 implementation
from itertools import combinations

def find(lst, n):
    print('YES' if [1 for r in range(1, len(lst) + 1)
                      for subset in combinations(lst, r)
                      if sum(subset) ==   n] else 'NO')

find([-1, 2, 4, 121], 5)

#This code is contributed by signi dimitri

C#

// C# implementation of the above approach
using System;
public class GFG
{

        // Find way to sum to N using array elements atmost once
        static void find(int [] arr, int length, int s)
        {
            // loop for all 2^n combinations
            for (int i = 1; i <= (Math.Pow(2, length)); i++) {

                // sum of a combination
                int sum = 0;

                for (int j = 0; j < length; j++)

                    // if the bit is 1 then add the element
                    if (((i >> j) & 1) % 2 == 1)
                        sum += arr[j];

                // if the sum is equal to given sum print yes
                if (sum == s) {
                    Console.Write("YES");
                    return;
                }
            }

            // else print no
            Console.Write("NO");
        }

        // driver code
        public static void Main()
        {
            int sum = 5;
            int []array = { -1, 2, 4, 121 };
            int length = array.Length;

            // find whether it is possible to sum to n
            find(array, length, sum);

        }

}

// This code is contributed by PrinciRaj19992

PHP

<?php
// PHP implementation of the above approach

// Find way to sum to N using array elements atmost once

function find($arr, $length, $s)
{
    // loop for all 2^n combinations
    for ( $i = 1; $i < (pow(2, $length)); $i++)
    {

        // sum of a combination
        $sum = 0;

        for ($j = 0; $j < $length; $j++)

            // if the bit is 1 then add the element
            if ((($i >> $j) & 1))
                $sum += $arr[$j];

        // if the sum is equal to given sum print yes
        if ($sum == $s)
        {
            echo "YES","\n";
            return;
        }
    }

    // else print no
    echo "NO","\n";
}

    // Driver code
    $sum = 5;
    $array = array(-1, 2, 4, 121 );
    $length = sizeof($array) / sizeof($array[0]);

    // find whether it is possible to sum to n
    find($array, $length, $sum);


// This code is contributed by ajit.
?>

Javascript

<script>

// Javascript implementation of the above approach

// Find way to sum to N using array
// elements atmost once
function find(arr, length, s)
{
    // loop for all 2^n combinations
    for (var i = 1; i < (Math.pow(2, length)); i++)
    {

        // sum of a combination
        var sum = 0;

        for (var j = 0; j < length; j++)

            // if the bit is 1 then add the element
            if (((i >> j) & 1))
                sum += arr[j];

        // if the sum is equal to given sum print yes
        if (sum == s) {
            document.write( "YES" + "<br>");
            return;
        }
    }

    // else print no
    document.write( "NO" + "<br>");
}


var sum = 5;
var array = [ -1, 2, 4, 121 ];
var length = array.length;

// find whether it is possible to sum to n
find(array, length, sum);

// This code is contributed by SoumikMondal

</script>

Output:

YES

Note: This program would not run for the large size of the array.