3.cpp

/*

Two City Scheduling
-------------------

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

 

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.
 

Note:

1 <= costs.length <= 100
It is guaranteed that costs.length is even.
1 <= costs[i][0], costs[i][1] <= 1000
*/

auto cmp = [](const vector<int>& lhs, const vector<int>& rhs) {
    return lhs[0] - lhs[1] < rhs[0] - rhs[1];
};

class Solution {
public:
    
    int twoCitySchedCost(vector<vector<int>>& costs) {
        long long sum = 0;
        sort(costs.begin(), costs.end(), cmp);
        int n = costs.size()/2;
        for(int i=0; i<2*n; i++) {
            if(i < n) {
                sum += costs[i][0];
            } else {
                sum += costs[i][1];
            }
        }
        return sum;
    }
};