Subset array sum by generating all the subsets
Last Updated :
04 Feb, 2022
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
#include <bits/stdc++.h>
using namespace std;
void find( int arr[], int length, int s)
{
for ( int i = 1; i < ( pow (2, length)); i++) {
int sum = 0;
for ( int j = 0; j < length; j++)
if (((i >> j) & 1))
sum += arr[j];
if (sum == s) {
cout << "YES" << endl;
return ;
}
}
cout << "NO" << endl;
}
int main()
{
int sum = 5;
int array[] = { -1, 2, 4, 121 };
int length = sizeof (array) / sizeof ( int );
find(array, length, sum);
return 0;
}
|
Java
class GFG
{
static void find( int [] arr, int length, int s)
{
for ( int i = 1 ; i <= (Math.pow( 2 , length)); i++) {
int sum = 0 ;
for ( int j = 0 ; j < length; j++)
if (((i >> j) & 1 ) % 2 == 1 )
sum += arr[j];
if (sum == s) {
System.out.println( "YES" );
return ;
}
}
System.out.println( "NO" );
}
public static void main(String[] args)
{
int sum = 5 ;
int []array = { - 1 , 2 , 4 , 121 };
int length = array.length;
find(array, length, sum);
}
}
|
Python3
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 )
|
C#
using System;
public class GFG
{
static void find( int [] arr, int length, int s)
{
for ( int i = 1; i <= (Math.Pow(2, length)); i++) {
int sum = 0;
for ( int j = 0; j < length; j++)
if (((i >> j) & 1) % 2 == 1)
sum += arr[j];
if (sum == s) {
Console.Write( "YES" );
return ;
}
}
Console.Write( "NO" );
}
public static void Main()
{
int sum = 5;
int []array = { -1, 2, 4, 121 };
int length = array.Length;
find(array, length, sum);
}
}
|
PHP
<?php
function find( $arr , $length , $s )
{
for ( $i = 1; $i < (pow(2, $length )); $i ++)
{
$sum = 0;
for ( $j = 0; $j < $length ; $j ++)
if ((( $i >> $j ) & 1))
$sum += $arr [ $j ];
if ( $sum == $s )
{
echo "YES" , "\n" ;
return ;
}
}
echo "NO" , "\n" ;
}
$sum = 5;
$array = array (-1, 2, 4, 121 );
$length = sizeof( $array ) / sizeof( $array [0]);
find( $array , $length , $sum );
?>
|
Javascript
<script>
function find(arr, length, s)
{
for ( var i = 1; i < (Math.pow(2, length)); i++)
{
var sum = 0;
for ( var j = 0; j < length; j++)
if (((i >> j) & 1))
sum += arr[j];
if (sum == s) {
document.write( "YES" + "<br>" );
return ;
}
}
document.write( "NO" + "<br>" );
}
var sum = 5;
var array = [ -1, 2, 4, 121 ];
var length = array.length;
find(array, length, sum);
</script>
|
Note: This program would not run for the large size of the array.
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