We have a set of items: the i-th item has value values[i] and label labels[i].
Then, we choose a subset S of these items, such that:
|S| <= num_wanted
For every label L, the number of items in S with label L is <= use_limit.
Return the largest possible sum of the subset S.
Example 1:
Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], num_wanted = 3, use_limit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth item.
Example 2:
Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], num_wanted = 3, use_limit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third item.
Example 3:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 1
Output: 16
Explanation: The subset chosen is the first and fourth item.
Example 4:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 2
Output: 24
Explanation: The subset chosen is the first, second, and fourth item.
Note:
1 <= values.length == labels.length <= 20000
0 <= values[i], labels[i] <= 20000
1 <= num_wanted, use_limit <= values.length
Solution:
class Solution {
public int largestValsFromLabels(int[] values, int[] labels, int num_wanted, int use_limit) {
int n = values.length;
int[][] nodes = new int[n][2];
for (int i = 0; i < n; i ++) {
nodes[i] = new int[]{values[i], labels[i]};
}
Arrays.sort(nodes, (a, b) -> b[0] - a[0]);
int count = 0, sum = 0;
int[] map = new int[20001];
for (int i = 0; i < n; i ++) {
int[] node = nodes[i];
if (map[node[1]] < use_limit) {
sum += node[0];
map[node[1]] ++;
count ++;
}
if (count == num_wanted) break;
}
return sum;
}
}