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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. 1 <= values.length == labels.length <= 20000
  2. 0 <= values[i], labels[i] <= 20000
  3. 1 <= num_wanted, use_limit <= values.length

Related Topics:
Hash Table, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/largest-values-from-labels/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int largestValsFromLabels(vector<int>& values, vector<int>& labels, int num_wanted, int use_limit) {
        int N = values.size();
        vector<int> id(N);
        iota(begin(id), end(id), 0);
        sort(begin(id), end(id), [&](int a, int b) { return values[a] > values[b]; });
        unordered_map<int, int> m;
        int ans = 0;
        for (int i = 0; i < N && num_wanted > 0; ++i) {
            int j = id[i];
            if (m[labels[j]] >= use_limit) continue;
            ans += values[j];
            m[labels[j]]++;
            num_wanted--;
        }
        return ans;
    }
};