LeetCode 3188: Find Top Scoring Students II Solution

Master LeetCode problem 3188 (Find Top Scoring Students II), 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.

3188. Find Top Scoring Students II

Hard

Database

Problem Explanation

Explanation:

To find the top scoring students II, we can use a priority queue to keep track of the top k students based on their scores. We will iterate through the students' scores and add them to the priority queue. If the size of the priority queue exceeds k, we will remove the lowest scoring student from the queue. At the end, we will have the top k scoring students in the priority queue.

Time Complexity:

The time complexity of this approach is O(n log k), where n is the number of students and k is the number of top scoring students we want to find.

Space Complexity:

The space complexity of this approach is O(k) to store the top scoring students in the priority queue.

: :

Solution Code

import java.util.PriorityQueue;

class Solution {
    public int[] findTopScoringStudents(int[] scores, int k) {
        PriorityQueue<Integer> pq = new PriorityQueue<>(k);
        
        for (int score : scores) {
            pq.offer(score);
            if (pq.size() > k) {
                pq.poll();
            }
        }
        
        int[] topScores = new int[k];
        for (int i = k - 1; i >= 0; i--) {
            topScores[i] = pq.poll();
        }
        
        return topScores;
    }
}

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What is the time complexity of LeetCode 3188 (Find Top Scoring Students II)?

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