Two City Scheduling — LeetCode 1029 Python Solution

The problem

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.

Example

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.

Python solution

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        costs.sort(key=lambda x: x[0] - x[1])
        n = len(costs) >> 1
        return sum(costs[i][0] + costs[i + n][1] for i in range(n))

Complexity

Pattern: Greedy

Take the locally best option every time — when you can prove that never costs you later. LeetCode 1029. Two City Scheduling is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Greedy.

The greedy guide has the Python template for the pattern and the 346 LeetCode problems that use it.

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