Two City Scheduling

LeetCode Q 1029 - Two City Scheduling

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

Solution

Code:

public int twoCitySchedCost(int[][] costs) {
	
	Arrays.sort(costs, new Comparator<int[]>() {
		public int compare (int[] a, int[] b) {
			return (a[1] - a[0]) - (b[1] - b[0]);
		}	
	});

	int cost = 0;

	for (int i = 0; i < costs.length / 2; i++) 
		cost += (costs[i][1] + costs[costs.length - 1 - i][0]);

	return cost;
}