## 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 - v).collect();
diffs.sort_unstable();
let sum_of_b: i32 = costs.iter().map(|v| v).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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