1029.TwoCitySchedCost.cs

// 1029. Two City Scheduling
// 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.

// Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.
// Example 1:
// 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.
// 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.
// Example 2:
// Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
// Output: 1859
// Example 3:
// Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
// Output: 3086
// Constraints:
// 2 * n == costs.length
// 2 <= costs.length <= 100
// costs.length is even.
// 1 <= aCosti, bCosti <= 1000

public class Solution {
    public int TwoCitySchedCost(int[][] costs) {
        //Sort on the basis of how much more I will be profited if I choose city 1 insted of city 2;
        costs = costs.OrderBy(x => x[0]-x[1]).ToArray(); 
        int cost = 0;
        for(int i = 0; i < costs.Length; i++){
            if(i<costs.Length/2)
                cost += costs[i][0];
            else
                cost += costs[i][1];
        }
        return cost;
    }
}