LeetCode 618: Students Report By Geography Solution

Master LeetCode problem 618 (Students Report By Geography), 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.

618. Students Report By Geography

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a binary search approach. We can start by sorting the given list of student scores. Then, for each query, we can perform a binary search to find the number of students with scores between the given range.

Here are the steps:

Sort the list of student scores in ascending order.
For each query, use binary search to find the index of the first student with a score greater than or equal to the lower bound, and the index of the last student with a score less than or equal to the upper bound.
Calculate the number of students within the range by subtracting the indexes found in step 2 and adding 1.

Time complexity analysis:

Sorting the list of student scores takes O(nlogn) time.
Processing each query using binary search takes O(logn) time.
Therefore, the overall time complexity is O(nlogn + qlogn), where n is the number of students and q is the number of queries.

Space complexity:

We only need a constant amount of extra space for variables, so the space complexity is O(1).

Solution Code

class Solution {
    public List<Integer> countStudents(int[] heights, int[][] people) {
        Arrays.sort(heights);
        List<Integer> result = new ArrayList<>();
        
        for (int[] p : people) {
            int low = 0, high = heights.length - 1;
            int start = -1, end = -1;
            
            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (heights[mid] >= p[0]) {
                    start = mid;
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            
            low = 0;
            high = heights.length - 1;
            
            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (heights[mid] <= p[1]) {
                    end = mid;
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
            
            result.add(end - start + 1);
        }
        
        return result;
    }
}

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