1029.two-city-scheduling.cs

/*
 * @lc app=leetcode id=1029 lang=csharp
 *
 * [1029] Two City Scheduling
 *
 * https://leetcode.com/problems/two-city-scheduling/description/
 *
 * algorithms
 * Easy (56.46%)
 * Likes:    785
 * Dislikes: 106
 * Total Accepted:    32.7K
 * Total Submissions: 57.9K
 * Testcase Example:  '[[10,20],[30,200],[400,50],[30,20]]'
 *
 * 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
 *
 */
using System.Linq;
namespace _1029
{

    // @lc code=start
    public class Solution
    {
        // https://leetcode.com/problems/two-city-scheduling/discuss/278898/Java-2ms-sorting-solution-with-explanation
        public int TwoCitySchedCost(int[][] costs) =>
            costs.OrderBy(cost => cost[0] - cost[1])
                .Select((cost, i) => i < costs.Length /2 ?  cost[0] : cost[1])
                .Sum();
    }
    // @lc code=end
}