H-Index - Google Top Interview Questions


Problem Statement :


In academia, the h-index is a metric used to calculate the impact of a researcher's papers. It is calculated as follows:

A researcher has index h if at least h papers have at least h citations each. If there are multiple h satisfying this formula, the maximum is chosen.

Given a list of paper citations of a researcher citations, calculate their h-index.

Constraints

n ≤ 100,000 where n is the length of citations
Example 1
Input
citations = [4, 3, 0, 1, 5]
Output
3
Explanation
At least 3 papers have at least 3 citations each: 3, 4, 5.

Example 2
Input
citations = [4, 4, 0, 1, 5, 9]
Output
4
Explanation
At least 4 papers have 4 citations each: 4, 4, 5, 9.

Example 3


Input

citations = [9, 10, 0, 1, 6]

Output

3

Explanation

At least 3 papers have 3 citations each: 6, 9, 10.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& citations) {
    int h = 0, N = citations.size();
    sort(citations.begin(), citations.end());
    for (int i = 0; i < N; i++) {
        if (citations[i] >= h && N - i >= h) {
            h = min(citations[i], N - i);
        }
    }
    return h;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] citations) {
        Arrays.sort(citations);
        int n = citations.length;
        int hIndex = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            int hPapers = n - i;
            int hCitations = citations[i];
            hIndex = Math.max(hIndex, Math.min(hPapers, hCitations));
        }
        return hIndex;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, citations):
        lo, hi = 0, len(citations)
        while lo < hi:
            mid = (lo + hi + 1) // 2
            cnt = 0
            for citation in citations:
                if citation >= mid:
                    cnt += 1
            if cnt >= mid:
                lo = mid
            else:
                hi = mid - 1
        return lo


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