1029-Two-City-Scheduling.py

'''
There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.



Example 1:

Input: [[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.


Note:

1 <= costs.length <= 100
It is guaranteed that costs.length is even.
1 <= costs[i][0], costs[i][1] <= 1000
'''
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        costs.sort(key = lambda x: x[1]-x[0])

        res, first_half = 0, len(costs)//2

        for i in range(len(costs)):
            res+=costs[i][1] if i < first_half else costs[i][0]

        return res