algorithms/cpp/h-Index/h-Index.cpp

// Source : https://leetcode.com/problems/h_index/
// Author : Calinescu Valentin
// 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 <algorithm>
class Solution {
public:
    int hIndex(vector<int>& 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;
    }
};