Coins(Binary Search - HackerEarth)

/*COINS
There are N bags and each bag contains some coin(s). Your task is to select an integer X and remove all the bags in which the number of coins is equal to X. Divide the remaining bags into two non-empty groups such that:

 The number of coin(s) in each bag of the first group is strictly smaller than X.
The number of coin(s) in each bag of the second group is strictly larger than X.
The total number of coins of one group is equal to the other.
Input Format:

The first line contains an integer N denoting the number of bags.
The second line contains N space-separated integers,A[i] denoting the number of coins in ith  bags. The  integer denotes the values of A[i](1-10^5).

Output Format:

Print ,YES if it is possible to divide the bags into two groups, else print NO.
*/
#include<bits/stdc++.h>
 
using namespace std;
typedef long long int lli;

int main()
{
    int n;
    cin >> n;
    int a[n];
    for(auto &x: a) cin >> x;
    int lo = 1, hi = 100000, mid;
    while(lo <= hi){
        mid = (lo+hi)/2;
        lli small=0, big=0;
        for(int i=0; i<n; i++){
            if(a[i] > mid){
                big += a[i];
            }
            else if(a[i] < mid){
                small += a[i];
            }
        }
        if(big != 0 && big == small){
            printf("YES\n");
            return 0;
        }
        if(big > small){
            lo = mid+1;
        }
        else{
            hi = mid-1;
        }
    }
    printf("NO\n");
 
    return 0;
}