1029. Two City Scheduling

題目敘述

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

題目翻譯

今天公司需要面試 2n 個人，但因為人太多所以需要分散到兩個城市 A, B 來面試。有一個陣列 costs 儲存了第 i 個人分別到 A, B 城市的花費。costs[0] = [40, 20]，代表第一個人到 A 城市要 40，到 B 城市要 20。然後要計算出分配後最少公司需要花費多少錢。

解法解析

這題很簡單的想法，我要最少的花費那當然就是每個人的花費中取最小的就好了。但是這樣就會有個盲點是，不要忘記了還要是平均分攤兩個城市，不然的話像是 Example2：259 A, 54 B, 667 B, 139 B, 118 B, 469 B，就會發現 B 太多了，所以排序的條件需要變成使用價差的方式 aCost - bCost。然後將排序後的結果裁半，一半到 A 一半到 B。

解法範例

Go

func twoCitySchedCost(costs [][]int) int {
    sort.Slice(costs, func(i, j int) bool {
        return costs[i][0]-costs[i][1] < costs[j][0]-costs[j][1]
    })
    var total int
    n := len(costs) / 2
    for i := 0; i < n; i++ {
        total += costs[i][0] + costs[i+n][1]
    }
    return total
}

JavaScript

/**
 * @param {number[][]} costs
 * @return {number}
 */
var twoCitySchedCost = function (costs) {
    costs.sort((a, b) => a[0] - a[1] - (b[0] - b[1]));
    let total = 0;
    const n = costs.length / 2;
    for (let i = 0; i < n; i++) {
        total += costs[i][0] + costs[i + n][1];
    }
    return total;
};

Kotlin

class Solution {
    fun twoCitySchedCost(costs: Array<IntArray>): Int {
        costs.sortWith(compareBy { it[0] - it[1] })

        val n = costs.size / 2
        return costsByDiffs.withIndex().sumBy {
            it.value[ if (it.index < n) 0 else 1 ]
        }
    }
}

PHP

class Solution
{

    /**
     * @param Integer[][] $costs
     * @return Integer
     */
    function twoCitySchedCost($costs)
    {
        usort($costs, function ($a, $b) {
            return ($a[0] - $a[1]) - ($b[0] - $b[1]);
        });

        $total = 0;
        $n = count($costs) / 2;
        for ($i = 0; $i < $n; $i++) {
            $total += $costs[$i][0] + $costs[$i + $n][1];
        }
        return $total;
    }
}

Python

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        # Sort by a gain which company has
        # by sending a person to city A and not to city B
        costs.sort(key=lambda x: x[0] - x[1])

        total = 0
        n = len(costs) // 2
        # To optimize the company expenses,
        # send the first n persons to the city A
        # and the others to the city B
        for i in range(n):
            total += costs[i][0] + costs[i + n][1]
        return total

Rust

impl Solution {
    pub fn two_city_sched_cost(costs: Vec<Vec<i32>>) -> i32 {
        let mut costs = costs;
        costs.sort_by(|a, b| (a[0] - a[1]).cmp(&(b[0] - b[1])));
        let mut total = 0;
        let n: usize = costs.len() / 2;
        for i in 0..n {
            total += costs[i][0] + costs[i + n][1];
        }
        total
    }
}

Swift

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        var costs = costs
        costs.sort { $0[0] - $0[1] < $1[0] - $1[1] }
        var total = 0
        let n = costs.count / 2
        for i in 0..<n {
            total += costs[i][0] + costs[i + n][1]
        }
        return total
    }
}