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LeetCode 1092 Two City Scheduling - Medium

1029. Two City Scheduling -- Medium

·1 min read·#LeetCode#Medium#Python

1092. Two City Scheduling — Medium

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Problem

  1. Two City Scheduling -- Medium

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: 2n == costs.length 2 <= costs.length <= 100 costs.length is even. 1 <= aCosti, bCosti <= 1000

Solution

### Greedy:
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        # Sort by a gain which company has 
        # by sending a person to city A and not to city B
        costs.sort(key = lambda x : x[0] - x[1])
        total = 0
        n = len(costs) // 2
        # To optimize the company expenses,
        # send the first n persons to the city A
        # and the others to the city B
        for i in range(n):
            total += costs[i][0] + costs[i + n][1]
        return total

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