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:
1
2
3
4
5
6
7
8
9
| 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
Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| class Solution {
public int twoCitySchedCost(int[][] costs) {
int sum = 0;
int n = costs.length;
Arrays.sort(costs, (a, b) -> {
return Integer.compare(a[0] - a[1], b[0] - b[1]);
});
for (int i = 0; i < n; i++) {
if (i >= n/2) {
sum += costs[i][1];
} else {
sum += costs[i][0];
}
}
return sum;
}
}
|