In this HackerEarth Sum of Numbers problem solution, we have given an array of N elements, check if it is possible to obtain a sum of S, by choosing some (or none) elements of the array and adding them.


HackerEarth Sum of Numbers problem solution


HackerEarth Sum of Numbers problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll arr[20];
void combinationUtil(ll arr[],ll n,ll r,ll index,ll data[],ll i);
void printCombination(ll arr[], ll n, ll r)
{
    ll data[r];
    combinationUtil(arr, n, r, 0, data, 0);
}
ll ans,flag;
void combinationUtil(ll arr[], ll n, ll r, ll index, ll data[], ll i)
{
    if(index == r)
    {
        ll sum=0;
        for (ll j=0; j<r; j++)//sum up all the items being included in the set
            sum+=data[j];
        if(sum==ans)//check against sum 
            flag=1;
        return;
    }
    if (i >= n)//if number of elements exceeds N
        return;
    data[index] = arr[i];//set element
    combinationUtil(arr, n, r, index+1, data, i+1);//item is being included
    combinationUtil(arr, n, r, index, data, i+1);//item is not included
}
int main()
{
    ll a,b,i,j,num,loop_sum;
    int test;
    scanf("%d",&test);
    while(test--)
    {
        flag=0;
        scanf("%lld",&num);
        for(i=0;i<num;i++)
            scanf("%lld",&arr[i]);
        scanf("%lld",&ans);
        for(ll r=0;r<=num;r++)
            printCombination(arr,num,r);
        if(flag==1)
            printf("YES\n");
        else
            printf("NO\n");
    }
}

Second solution

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;cin>>t;
    while(t--){
        int n; cin>>n; int a[n+5];
        for(int i=0;i<n;i++) cin>>a[i];
        int s,i,j; cin>>s;
        for(i=0;i<(1<<n);i++){
            int x=0;
            for(j=0;j<n;j++){
                if(i & (1<<j)) x+=a[j];
            }
            if(x==s){
                cout<<"YES\n"; break;
            }
        }
        if(i==(1<<n)) {cout<<"NO\n";} 
    }
    return 0;
}