HackerEarth Median Game problem solution

YASH PAL, 31 July 202415 February 2026

In this HackerEarth Median Game problem solution, You are given an array A of N integers. You perform this operation N – 2 times: For each contiguous subarray of odd size greater than 2, you find the median of each subarray(Say medians obtained in a move are m1,m2,m3,..,mk).

In each move, you remove the first occurrence of value min(m1,m2,m3,…,mk) from the original array. After removing the element the array size reduces by 1 and no void spaces are left. For example, if we remove element 2 from the array {1,2,3,4}, the new array will be {1,3,4}.

Print a single integer denoting the sum of numbers that are left in the array after performing the operations. You need to do this for T test cases.

HackerEarth Median Game problem solution

HackerEarth Median Game problem solution.

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

#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL);

int n, t;

int main() {
    IOS;
    cin>>t;
    while(t--){
        cin>>n;
        int mn = INT_MAX, mx = -1;
        for(int i = 0; i < n; i++){
            int u;
            cin>>u;
            mx = max(mx, u);
            mn = min(mn, u);
        }
        cout<<mx + mn<<"n";
    }
    return 0;
}

Second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5;
int main(){
   ios::sync_with_stdio(0), cin.tie(0);
   int t;
   cin >> t;
   while(t--){
      int n;
      cin >> n;
      int mn = INT_MAX, mx = -mn;
      while(n--){
         int x;
         cin >> x;
         mn = min(mn, x);
         mx = max(mx, x);
      }
      cout << mn + mx << 'n';
   }
}

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