H-Index II – 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 and citations is sorted in an ascending order, 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.

You must write an algorithm that runs in logarithmic time.

Example 1:

Input: citations = [0,1,3,5,6]
Output: 3
Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had received 0, 1, 3, 5, 6 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 = [1,2,100]
Output: 2

Constraints:

• n == citations.length
• 1 <= n <= 105
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

C++ H-Index II LeetCode Solution

int hIndex(vector<int>& c) {
        int n = c.size();
        if(!n) return 0;
        for(int i=0;i<n;i++){
             if(c[i] >= n-i) return n-i;   // The first element whose value is more than the length of remaining array. 
			 //So we return the remaining length which is the answer.
			 // eg [0,1,3,4,6]  c[2] = 3 >2(length of remaing array) so n-i = length of remaining array + that element
        }
        return 0;
    }

Java H-Index II LeetCode Solution

public int hIndex(int[] citations) {
	int len = citations.length;
	int lo = 0, hi = len - 1;
	while (lo <= hi) {
		int med = (hi + lo) / 2;
		if (citations[med] == len - med) {
			return len - med;
		} else if (citations[med] < len - med) {
			lo = med + 1;
		} else { 
			//(citations[med] > len-med), med qualified as a hIndex,
		    // but we have to continue to search for a higher one.
			hi = med - 1;
		}
	}
	return len - lo;
}
 

Python 3 H-Index II LeetCode Solution

class Solution:
    def hIndex(self, citations):
        if not citations: return 0
        n = len(citations)
        beg, end = 0, n - 1
        while beg <= end:
            mid = (beg + end)//2
            if mid + citations[mid] >= n:
                end = mid - 1
            else:
                beg = mid + 1                
        return n - beg

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