# Two City Schedulingš¦¦

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:110Explanation:

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`

**Solution**

Time complexity

O(n log n)Space complexity

O(1)