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:
        a = 0
        b = 0
        cost = 0
        costs.sort(key=lambda x : (abs(x[0]-x[1]),x[0]), reverse=True)
        print(costs)
        for i in range(len(costs)):
            if costs[i][0]<=costs[i][1]:
                if a<len(costs)//2:
                    cost+=costs[i][0]
                    a+=1
                else:
                    cost+=costs[i][1]
                    b+=1
            else:
                if b<len(costs)//2:
                    cost+=costs[i][1]
                    b+=1
                else:
                    cost+=costs[i][0]
                    a+=1
                
        return cost