Two City Scheduling

#leetcode #dsa #theabbie

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

SOLUTION:

from functools import cmp_to_key

class Solution:
    def compare(self, cost1, cost2):
        a, b = cost1
        c, d = cost2
        if a + d <= b + c:
            return -1
        return 1

    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        n = len(costs)
        costs.sort(key = cmp_to_key(self.compare))
        cost = 0
        for i in range(n // 2):
            cost += costs[i][0]
        for i in range(n // 2, n):
            cost += costs[i][1]
        return cost