Two City Scheduling

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

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

Example 1:

Input: costs = [[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.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Constraints:

2n == costs.length
2 <= costs.length <= 100
costs.length is even.
1 <= aCosti, bCosti <= 1000

Solution:

class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int n = costs.length / 2;
        Arrays.sort(costs, (a, b) -> Integer.compare(b[1] - b[0], a[1] - a[0]));
        int a = 0;
        int total = 0;
        for (int[] c : costs) {
            if (a < n) {
                a ++;
                total += c[0];
            } else {
                total += c[1];
            }
            // System.out.print(Arrays.toString(c));
        }
        // System.out.println();
        return total;
    }
}