LeetCode - Two City Scheduling

問題の説明

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.

I/O例

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

せいげんじょうけん

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

Solution

[戦略]
1、2 n人とも都会Bに行った時の和.
2.(Aが選択されて増加した費用-Bが選択されずに減少した費用)がpriorityqueueを構成する.
3、総和最小を方向として、n人を変えてA市を選ぶ.

import java.util.PriorityQueue;

class Solution {
    public int twoCitySchedCost(int[][] costs) {

        int minSumB = 0;
        for (int i = 0; i < costs.length; i++) {
            minSumB += costs[i][1];
        }

        PriorityQueue<Integer> abDiff = new PriorityQueue<>();

        for (int i = 0; i < costs.length; i++) {
            abDiff.add(costs[i][0] - costs[i][1]);
        }

        for (int i = 0; i < costs.length / 2; i++) {
            minSumB += abDiff.poll();
        }
        return minSumB;

    }

}