LeetCode [1029] Two City Scheduling

 1029. Two City Scheduling

Medium

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:

• 2n == costs.length
• 2 <= costs.length <= 100
• costs.length is even.
• 1 <= aCosti, bCosti <= 1000

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class Solution { public int twoCitySchedCost(int[][] costs) { int n = costs.length; int[] diff = new int[n]; int minCost = 0, index = 0; for(int[] cost : costs){ diff[index++] = cost[1] - cost[0]; minCost += cost[0]; } Arrays.sort(diff); for(int i = 0; i < n/2; i++){ minCost += diff[i]; } return minCost; } }