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H-Index LeetCode Solution

Problem – H-Index LeetCode Solution

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return compute the researcher’s h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

Example 1:

Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 citations respectively.
Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3 citations each, their h-index is 3.

Example 2:

Input: citations = [1,3,1]
Output: 1


  • n == citations.length
  • 1 <= n <= 5000
  • 0 <= citations[i] <= 1000

H-Index LeetCode Solution in Java

public int hIndex(int[] citations) {
    int n = citations.length;
    int[] buckets = new int[n+1];
    for(int c : citations) {
        if(c >= n) {
        } else {
    int count = 0;
    for(int i = n; i >= 0; i--) {
        count += buckets[i];
        if(count >= i) {
            return i;
    return 0;

H-Index LeetCode Solution in Python

    return sum(i < j for i, j in enumerate(sorted(citations, reverse=True)))

H-Index LeetCode Solution in C++

class Solution {
    int hIndex(vector<int>& c) {
        // 3 support variables for us
        int s = 0, e = c.size() - 1, avg;
        // the base of every happy binary search ever: having a sorted dataset
        sort(begin(c), end(c));
        // some good old binary search here to find the maximum element meeting the conditions
        while (s <= e) {
            if (c[avg = (e + s) / 2] < c.size() - avg) s = avg + 1;
            else e = avg - 1;
        return c.size() - s;
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