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 oncevoid 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 codeint 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 approachclass 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 implementationfrom itertools import combinationsdef 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 approachusing 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 oncefunction 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 oncefunction 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 nfind(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.
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