README.md

1029. 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. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

Solutions (Python)

1. Solution

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        N = len(costs) / 2
        A, B = [], []
        costs.sort(key=lambda cost : abs(cost[0] - cost[1]), reverse=True)
        for cost in costs:
            if (cost[0] <= cost[1] and len(A) < N) or len(B) == N:
                A.append(cost)
            elif (cost[0] >= cost[1] and len(B) < N) or len(A) == N:
                B.append(cost)
        return sum(cost[0] for cost in A) + sum(cost[1] for cost in B)