# 1029. Two City Scheduling - June day 3

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.``````

To optimize the total costs, let's sort the persons by priceA - priceB and then send the first n persons to the city A, and the others to the city B, because this way the company costs are minimal.

``````class Solution {

public int twoCitySchedCost(int[][] costs) {
Arrays.sort(costs, (a, b) -> {
return (a[0] - a[1]) - (b[0] - b[1]);
});

var total = 0;
var n = costs.length / 2;

for (int i = 0; i < n; ++i) {
total += costs[i][0];
total += costs[i + n][1];
}