# 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`
 ``````1 2 3 4 5 6 7 8 9 `````` ``````class Solution: def twoCitySchedCost(self, costs: List[List[int]]) -> int: # sort by sending a person to A and not to B costs.sort(key=lambda x: x[0] - x[1]) total = 0 n = len(costs) // 2 for i in range(n): total += costs[i][0] + costs[i + n][1] return total``````