1029. Two City Scheduling.cpp

#JuneChallenge
/*
Link: https://leetcode.com/problems/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
*/

// Solution

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        vector<int> refund;
        int N = costs.size()/2;
        int minCost = 0;
        for(vector<int> cost : costs){
            minCost += cost[0];
            refund.push_back(cost[1] - cost[0]); 
        }
        sort(refund.begin(), refund.end());
        for(int i = 0; i < N; i++){
            minCost += refund[i];
        }
        return minCost;
    }
};