LeetCode - Two City Scheduling

문제 설명

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.

입출력 예시

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

제약사항

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

Solution

[전략]
1. 2n명의 사람이 모두 city B로 갔을때의 합을 구한다.
2. (A를 선택해서 추가되는 비용 - B를 선택하지 않아서 감소되는 비용)으로 priorityqueue를 구성한다.
3. 전체 합이 최소화 되는 방향으로 n명이 city A를 선택하도록 변경한다.

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;

    }

}