1029. Two City Scheduling

Problem Statement

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

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 <= aCost_i, bCost_i <= 1000

Solution:

Rust

impl Solution {
    pub fn two_city_sched_cost(costs: Vec<Vec<i32>>) -> i32 {
        let mut costs = costs;

        let mut total = 0;
        let n = costs.len() / 2;

        // sort by a gain which company has
        // by sending a person to city A and not to city B
        costs.sort_by(|a, b| (a[0] - a[1]).cmp(&(b[0] - b[1])));

        // to optimize the company expenses,
        // send the first n persons to the city A
        // and the others to the city B
        for i in 0..n {
            total += costs[i][0] + costs[i + n][1]; // city A + city B
        }

        total
    }
}

1029. Two City Scheduling #

Problem Statement #

Solution: #

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1029. Two City Scheduling

Problem Statement

Solution:

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