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."

For example, given citations = [3, 0, 6, 1, 5], which 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, his h-index is 3.

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

Solution:

We can use a bucket sort like algorithm to solve this problem.

We use an frequency array with n + 1 length to record the frequency of citations.

If a citation is equal or greater than n, we increment frequency[n].

Then we go through frequency array reversely and check if the suffix sum at index i is equal or larger than i. If so, we know the number of citation of i papers is equal or larger than i, which is the h-index.

The time complexity is O(n) and the space complexity is also O(n).



Code:

public class Solution {
    public int hIndex(int[] citations) {
        if (citations == null || citations.length == 0) {
            return 0;
        }
        int n = citations.length;
        int[] frequency = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (citations[i] >= n) {
                frequency[n]++;
            }
            else {
                frequency[citations[i]]++;
            }
        }
        int count = 0;
        for (int i = n; i >= 0; i--) {
            count += frequency[i];
            if (count >= i) {
                return i;
            }
        }
        return 0;
    }
}