LeetCode in Kotlin

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

class Solution {
    fun twoCitySchedCost(costs: Array<IntArray>): Int {
        costs.sortWith { a: IntArray, b: IntArray ->
            a[0] - a[1] - (b[0] - b[1])
        }
        var cost = 0
        for (i in costs.indices) {
            cost += if (i < costs.size / 2) {
                costs[i][0]
            } else {
                costs[i][1]
            }
        }
        return cost
    }
}

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