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HIndex.java
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package array;
/**
* Created by gouthamvidyapradhan on 12/12/2017.
* 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 O(n) Replace all the citations which are greater than n with n, the result will not change with this
operation.
Maintain a count array with count of each citations. Sum up all the counts from n -> 0 and store this in a array S.
Value in array index Si is number of papers having citations at least i.
The first value at index i, from right to left in array S which has i <= Si is the answer.
*/
public class HIndex {
public static void main(String[] args) throws Exception{
int[] A = {3, 0, 6, 1, 5};
System.out.println(new HIndex().hIndex(A));
}
public int hIndex(int[] citations) {
int n = citations.length;
int[] count = new int[n + 1];
int[] S = new int[n + 1];
for(int i = 0; i < citations.length; i ++){
if(citations[i] > n){
citations[i] = n;
}
}
for (int citation : citations) {
count[citation]++;
}
S[n] = count[n];
for(int i = n - 1; i >= 0; i --){
S[i] = count[i] + S[i + 1];
}
for(int i = n; i >= 0; i--){
if(i <= S[i]){
return i;
}
}
return 0;
}
}