# H-Index II LeetCode Solution | Easy Approach Share:

H-Index II Given an array of integers `citations` where `citations[i]` is the number of citations a researcher received for their `ith` paper and `citations` is sorted in an ascending order, 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.

You must write an algorithm that runs in logarithmic time.

Example 1:

```Input: citations = [0,1,3,5,6]
Output: 3
Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had received 0, 1, 3, 5, 6 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 = [1,2,100]
Output: 2
```

Constraints:

• `n == citations.length`
• `1 <= n <= 105`
• `0 <= citations[i] <= 1000`
• `citations` is sorted in ascending order.

Time: O(log n)
Space: O(1)

### C++

``````class Solution {
public:
int hIndex(vector<int>& citations) {
int l = 0;
int r = citations.size();

while (l < r) {
const int m = l + (r - l) / 2;
if (citations[m] >= citations.size() - m)
r = m;
else
l = m + 1;
}

return citations.size() - l;
}
};
``````

### Java

``````
class Solution {
public int hIndex(int[] citations) {
int l = 0;
int r = citations.length;

while (l < r) {
final int m = l + (r - l) / 2;
if (citations[m] >= citations.length - m)
r = m;
else
l = m + 1;
}

return citations.length - l;
}
}
``````

### Python

``````  class Solution:
def hIndex(self, citations: List[int]) -> int:
l = 0
r = len(citations)

while l < r:
m = (l + r) // 2
if citations[m] >= len(citations) - m:
r = m
else:
l = m + 1

return len(citations) - l

``````

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