1029.Two_City_Scheduling.py

"""
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: 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.
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
"""

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        n = len(costs)
        res = 0
        q = []
        for i in range(n):
            res += costs[i][0]
            heapq.heappush(q, costs[i][1] - costs[i][0])
        for _ in range(n // 2):
            res += heapq.heappop(q)
        return res

       
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        # 排序标准：去city A比去city B多用多少钱
        # 这样一来去排在前面的就是去city A能省下最多钱的人
        # 让前N个人都去A就能省下最多的钱
        costs.sort(key = lambda x: (x[0] - x[1]))   
        N = len(costs) // 2
        total_cost = 0
        for i, [cost_A, cost_B] in enumerate(costs):
            total_cost = total_cost + cost_A if i < N else total_cost + cost_B
        return total_cost