# 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."
Example:
Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] 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, her h-index is 3.
Note: If there are several possible values for h, the maximum one is taken as the h-index.
# Solution
Approach 1: Sorting.
Approach 2: O(n).
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
// Sorting - ascending
class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
std::sort(citations.begin(), citations.end());
for (int i = 0; i < n; ++i)
{
if (citations[i] >= n - i)
return n - i;
}
return 0;
}
};
// Sorting - descending
class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
std::sort(citations.begin(), citations.end(), greater<int>());
for (int i = 0; i < n; ++i)
{
if (citations[i] >= i + 1 &&
(i == n - 1 || citations[i+1] <= i + 1))
return i + 1;
}
return 0;
}
};
Approach 2:
// O(n)
class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
vector<int> counter = vector<int>(n+1, 0);
for (int i = 0; i < n; ++i)
{
int j = citations[i];
if (j > n)
counter[n]++;
else
counter[j]++;
}
int total = 0;
for (int i = n; i >= 0; --i)
{
total += counter[i];
if (total >= i)
return i;
}
return 0;
}
};