LeetCode 2790: Maximum Number of Groups With Increasing Length Solution

Master LeetCode problem 2790 (Maximum Number of Groups With Increasing Length), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2790. Maximum Number of Groups With Increasing Length

Hard

Array Math Binary Search Greedy Sorting

Problem Explanation

Explanation

To solve this problem, we can use a greedy approach. We need to find the maximum number of groups we can create given the constraints. We iterate through the array usageLimits and for each element, we add it to a priority queue. The priority queue is sorted in descending order of available slots for each number. We then create groups by popping elements from the priority queue until the current element exceeds the number of available slots.

The time complexity of this approach is O(n log n), where n is the length of the usageLimits array, as we iterate through the array and perform operations on the priority queue. The space complexity is O(n) for the priority queue.

Solution Code

import java.util.PriorityQueue;

class Solution {
    public int maxGroupNumber(int[] usageLimits) {
        int n = usageLimits.length;
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> usageLimits[b] - usageLimits[a]);
        
        for (int i = 0; i < n; i++) {
            pq.offer(i);
        }
        
        int groups = 0;
        while (!pq.isEmpty()) {
            int curr = pq.poll();
            if (usageLimits[curr] == 0) continue;
            groups++;
            for (int i = curr + 1; i < n; i++) {
                if (usageLimits[i] > 0) {
                    usageLimits[i]--;
                    pq.offer(i);
                }
            }
        }
        
        return groups;
    }
}

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