1029-two-city-scheduling.py

"""
Problem Link: https://leetcode.com/problems/two-city-scheduling/

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