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.

`#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';   }}`