H-Index Given an array of integers citations
where citations[i]
is the number of citations a researcher received for their ith
paper, return compute the researcher’s h
-index.
According to the definition of h-index on Wikipedia: A scientist has an index h
if h
of their n
papers have at least h
citations each, and the other n − h
papers have no more than h
citations each.
If there are several possible values for h
, the maximum one is taken as the h
-index.
Example 1:
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, their h-index is 3.
Example 2:
Input: citations = [1,3,1] Output: 1
Constraints:
n == citations.length
1 <= n <= 5000
0 <= citations[i] <= 1000
H-Index Solutions
✅Time: O(n)
✅Space: O(n)
C++
class Solution {
public:
int hIndex(vector<int>& citations) {
const int n = citations.size();
int accumulate = 0;
vector<int> count(n + 1);
for (const int citation : citations)
++count[min(citation, n)];
// to find the largeset h-index, loop from back to front
// i is the candidate h-index
for (int i = n; i >= 0; --i) {
accumulate += count[i];
if (accumulate >= i)
return i;
}
throw;
}
};
Java
class Solution {
public int hIndex(int[] citations) {
final int n = citations.length;
int accumulate = 0;
int[] count = new int[n + 1];
for (final int citation : citations)
++count[Math.min(citation, n)];
// to find the largeset h-index, loop from back to front
// i is the candidate h-index
for (int i = n; i >= 0; --i) {
accumulate += count[i];
if (accumulate >= i)
return i;
}
throw new IllegalArgumentException();
}
}
Python
class Solution:
def hIndex(self, citations: List[int]) -> int:
n = len(citations)
accumulate = 0
count = [0] * (n + 1)
for citation in citations:
count[min(citation, n)] += 1
# to find the largeset h-index, loop from back to front
# i is the candidate h-index
for i, c in reversed(list(enumerate(count))):
accumulate += c
if accumulate >= i:
return i
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