// 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

//

// Runtime 1 ms Beats 99.2%

// Memory 40.6 MB Beats 76.84%

class Solution {

public int twoCitySchedCost(int[][] costs) {

int sum = 0;

int[] refunds = new int[costs.length];

for (int i = 0; i < costs.length; i++) {

sum = sum + costs[i][0];

refunds[i] = costs[i][0] - costs[i][1];

}

Arrays.sort(refunds);

for (int i = costs.length - 1; i >= costs.length / 2; i--) {

sum = sum - refunds[i];

}

return sum;

}

}