#LeetCode : Two City Scheduling

Also Check: June LeetCoding Challenge (2020)

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

GIT URL: Java Solution of Leet Code's Two City Scheduling problem
This problem is quite similar to the problem where we have to maximize profit. We can use greedy approach to solve this. Here the profit can be made by sending a person i to cityA or cityB. We need to find the absolute cost difference between two cities for all the person and then sort them descending order.

ALSO CHECK: Other LeetCode Solutions in Java

-K Himaanshu Shuklaa.