Day-03-Two City Scheduling.cpp

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




class Solution {
public:

   static bool comparator(vector<int> a, vector<int> b){
        return (abs(a[0]-a[1])) > (abs(b[0]-b[1]));
    }


    int twoCitySchedCost(vector<vector<int>>& costs) {
        int n=costs.size(), sum=0, ca=0, cb=0;
        sort(costs.begin(), costs.end(), comparator);
        for(auto cost: costs){
            if(cost[0]<cost[1] && ca<n/2){
                sum+=cost[0];  ca++;
            } else if(cost[1]<cost[0] && cb<n/2){
                sum+=cost[1];   cb++;
            } else {
                if(ca<n/2){ sum+=cost[0]; ca++;}
                else { sum+=cost[1];  cb++;}
            }
        }
        return sum;
    }
};