// Source : https://leetcode.com/problems/h_index/
// Author : Calinescu Valentin, Hao Chen
// Date : 2015-10-22
/***************************************************************************************
*
* 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.
*
***************************************************************************************/
/*
* Solutions
* =========
*
* A simple solution would be to sort the vector and then run through it starting with
* the last element. At every step we need to check whether this element is not less than
* the remaining number of elements bigger than it(including itself) and all the values of
* the other elements smaller than it are not more than that number. The h_index is this
* number of elements bigger than it(including itself).
*
* Time Complexity: O(N log N)
* Space Complexity: O(1)
*
*/
#include
class Solution {
public:
int hIndex(vector& citations) {
return hIndex02(citations);
return hIndex01(citations);
}
int hIndex01(vector& citations) {
sort(citations.begin(), citations.end());
int h_index = 0;
for(int i = citations.size() - 1; i >= 0; i--)
if(citations[i] >= citations.size() - i && (i - 1 < 0 || citations[i - 1] <= citations.size() - i))
h_index = citations.size() - i;
return h_index;
}
// same solution but a bit different implemtation
int hIndex02(vector& citations) {
sort(citations.begin(), citations.end());
int n = citations.size();
for (int i=0; i= n-i){
return n-i;
}
}
return 0;
}
};