1090. Largest Values From Labels

Problem

There is a set of n items. You are given two integer arrays values and labels where the value and the label of the ith element are values[i] and labels[i] respectively. You are also given two integers numWanted and useLimit.

Choose a subset s of the n elements such that:

• The size of the subset s is less than or equal to numWanted.

• There are at most useLimit items with the same label in s.

The score of a subset is the sum of the values in the subset.

Return **the maximum *score* of a subset **s.

  Example 1:

Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], numWanted = 3, useLimit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth items.

Example 2:

Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], numWanted = 3, useLimit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third items.

Example 3:

Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], numWanted = 3, useLimit = 1
Output: 16
Explanation: The subset chosen is the first and fourth items.

  Constraints:

• n == values.length == labels.length

• 1 <= n <= 2 * 10^4

• 0 <= values[i], labels[i] <= 2 * 10^4

• 1 <= numWanted, useLimit <= n

Solution (Java)

class Solution {
    private static class Node {
        int val;
        int label;

        Node(int val, int label) {
            this.val = val;
            this.label = label;
        }
    }

    public int largestValsFromLabels(int[] values, int[] labels, int numWanted, int useLimit) {
        PriorityQueue<Node> maxHeap =
                new PriorityQueue<>((a, b) -> b.val != a.val ? b.val - a.val : a.label - b.label);
        int n = values.length;
        for (int i = 0; i < n; i++) {
            maxHeap.offer(new Node(values[i], labels[i]));
        }
        int ans = 0;
        HashMap<Integer, Integer> labelAddedCount = new HashMap<>();
        while (!maxHeap.isEmpty() && numWanted > 0) {
            Node cur = maxHeap.poll();
            if (labelAddedCount.containsKey(cur.label)
                    && labelAddedCount.get(cur.label) >= useLimit) {
                continue;
            }
            if (cur.val > 0) {
                ans += cur.val;
                labelAddedCount.put(cur.label, labelAddedCount.getOrDefault(cur.label, 0) + 1);
                numWanted--;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).