# 274. H-Index

Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.

According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."

Example:

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, her h-index is 3.

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

# Solution

Approach 1: Sorting.

Approach 2: O(n).

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

// Sorting - ascending
class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        std::sort(citations.begin(), citations.end());
        for (int i = 0; i < n; ++i)
        {          
            if (citations[i] >= n - i)
                return n - i;
        }
        return 0;
    }
};

// Sorting - descending
class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        std::sort(citations.begin(), citations.end(), greater<int>());
        for (int i = 0; i < n; ++i)
        {          
            if (citations[i] >= i + 1 &&
                (i == n - 1 || citations[i+1] <= i + 1))
                return i + 1;
        }
        return 0;
    }
};

Approach 2:

// O(n)
class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        vector<int> counter = vector<int>(n+1, 0);
        for (int i = 0; i < n; ++i)
        {
            int j = citations[i];
            if (j > n)
                counter[n]++;
            else
                counter[j]++;
        }
        int total = 0;
        for (int i = n; i >= 0; --i)
        {
            total += counter[i];
            if (total >= i)
                return i;
        }
        return 0;
    }
};