Problem
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.length2 <= costs.length <= 100costs.lengthis even.1 <= aCosti, bCosti <= 1000
Solution (Java)
class Solution {
public int twoCitySchedCost(int[][] costs) {
Arrays.sort(costs, (a, b) -> (a[0] - a[1] - (b[0] - b[1])));
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;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).