# HackerEarth Pro and Con List problem solution

In this HackerEarth Pro and Con List problem solution, There is a long list of n girls in front of Barney, and he is to calculate the optimal "happiness" he can find by selecting exactly 2 girls. (Why 2? No one knows!)

Ted, as a fan of pros and cons, suggests to make a list, a method for estimating the maximum happiness that Barney can achieve.

Each girl is characterized by two parameters:

- favour: if this girl is chosen, his happiness increases by this amount.
- anger: if this girl is not chosen, his happiness decreases by this amount.

Find the maximum "happiness" that Barney can obtain. Note that the answer is allowed to be negative.

## HackerEarth Pro and Con List problem solution.

`#include <bits/stdc++.h>using namespace std;int main (){    int tc;        vector < int > favor;    vector < int > anger;    vector < long long int > ans;        scanf("%d",&tc);    assert(1<=tc);    assert(tc<=10);    while (tc--)    {        int n;        favor.clear();        anger.clear();        ans.clear();                scanf("%d",&n);        assert(2<=n);        assert(n<=100000);                favor.resize(n);        anger.resize(n);        ans.resize(n);                long long int hatao = 0;        for (int i=0; i<n; i++)        {            scanf("%d %d",&favor[i], &anger[i]);            ans[i] = favor[i] + anger[i];            hatao = hatao + anger[i];                        assert(0<=favor[i]);            assert(favor[i]<=1000000000);            assert(0<=anger[i]);            assert(anger[i]<=1000000000);        }        sort(ans.rbegin(),ans.rend());        cout << ans[0] + ans[1] - hatao << endl;    }    return 0;}`