# 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`

```python
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        n = len(costs) // 2
        # Tried sorting by A then B, but we need to sort actually by their diff
        # because that represents difference between going to A or B. 
        costs.sort(key=lambda x : x[0]-x[1])
        result = 0
        for A, B in costs[:n]:
            result += A
        for A, B in costs[n:]:
            result += B
        return result
            
```