LeetCode-in-Java

1029. Two City Scheduling

Medium

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:

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

Solution

import java.util.Arrays;

public class Solution {
    public int twoCitySchedCost(int[][] costs) {
        Arrays.sort(costs, (a, b) -> (a[0] - a[1] - (b[0] - b[1])));
        int cost = 0;
        for (int i = 0; i < costs.length; i++) {
            if (i < costs.length / 2) {
                cost += costs[i][0];
            } else {
                cost += costs[i][1];
            }
        }
        return cost;
    }
}

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