H-Index – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

Constraints:

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

C++  H-Index LeetCode Solution

class Solution {

public:
 
int hIndex(vector<int>& citations) {
    if(citations.empty())
        return 0;
    int n = citations.size();
    vector<int> hash(n + 1, 0);
    for(int i = 0; i < n; ++i){
        if(citations[i] >= n)
            hash[n]++;
        else
            hash[citations[i]]++;
    }
    int paper = 0;
    for(int i = n; i >= 0; --i){
        paper += hash[i];
        if(paper >= i)
            return i;
    }
}

 
};

Java  H-Index LeetCode Solution

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

Python 3  H-Index LeetCode Solution

def hIndex(self, citations):
    n = len(citations)
    citeCount = [0] * (n+1)
    for c in citations:
        if c >= n:
            citeCount[n] += 1
        else:
            citeCount[c] += 1
    
    i = n-1
    while i >= 0:
        citeCount[i] += citeCount[i+1]
        if citeCount[i+1] >= i+1:
            return i+1
        i -= 1
    return 0

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