package com.freetymekiyan.algorithms.level.medium;
/**
* 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.
* <p>
* 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."
* <p>
* 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.
* <p>
* Note: If there are several possible values for h, the maximum one is taken as the h-index.
* <p>
* Hint:
* <p>
* An easy approach is to sort the array first.
* What are the possible values of h-index?
* A faster approach is to use extra space.
* <p>
* Company Tags: Bloomberg, Google, Facebook
* Tags: Hash Table, Sort
* Similar Problems: (M) H-Index II
*/
public class HIndex {
/**
* Bucket sort.
* Suppose n is the number of papers.
* H can be at most n when a person has n papers and all of them have more than n citations.
* To find a number h that h of his n papers have >= h citations, put papers in buckets.
* All papers have >= n citations put into bucket n.
* Papers have i citations put into bucket i.
* Then count backwards.
* The first number i that has total papers >= i is the answer.
*/
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 papers = 0;
for (int i = n; i >= 0; i--) {
papers += buckets[i];
if (papers >= i) {
return i;
}
}
return 0;
}
}