two-city-scheduling

Two City Scheduling

LeetCode #: 1029

Difficulty: Easy

Topic: Greedy.

Problem

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

Complexity Analysis

Time complexity: O(n)

Space complexity: O(n)