1029. Two City Scheduling

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

Rust Solution

struct Solution;

impl Solution {
    fn two_city_sched_cost(costs: Vec<Vec<i32>>) -> i32 {
        let n = costs.len();
        let mut diffs: Vec<i32> = costs.iter().map(|v| v[0] - v[1]).collect();
        diffs.sort_unstable();
        let sum_of_b: i32 = costs.iter().map(|v| v[1]).sum();
        let sum_of_diff: i32 = diffs.iter().take(n / 2).sum();
        sum_of_b + sum_of_diff
    }
}

#[test]
fn test() {
    let costs: Vec<Vec<i32>> = vec_vec_i32![[10, 20], [30, 200], [400, 50], [30, 20]];
    assert_eq!(Solution::two_city_sched_cost(costs), 110);
}

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