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# H-Index II LeetCode Solution

## Problem – H-Index II LeetCode Solution

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.

## H-Index II LeetCode Solution in Java

``````public int hIndex(int[] citations) {
int len = citations.length;
int lo = 0, hi = len - 1;
while (lo <= hi) {
int med = (hi + lo) / 2;
if (citations[med] == len - med) {
return len - med;
} else if (citations[med] < len - med) {
lo = med + 1;
} else {
//(citations[med] > len-med), med qualified as a hIndex,
// but we have to continue to search for a higher one.
hi = med - 1;
}
}
return len - lo;
}
``````

## H-Index II LeetCode Solution in C++

``````int hIndex(vector<int>& c) {
int n = c.size();
if(!n) return 0;
for(int i=0;i<n;i++){
if(c[i] >= n-i) return n-i;   // The first element whose value is more than the length of remaining array.
//So we return the remaining length which is the answer.
// eg [0,1,3,4,6]  c = 3 >2(length of remaing array) so n-i = length of remaining array + that element
}
return 0;
}
``````

## H-Index II LeetCode Solution in Python

``````class Solution:
def hIndex(self, citations):
if not citations: return 0
n = len(citations)
beg, end = 0, n - 1
while beg <= end:
mid = (beg + end)//2
if mid + citations[mid] >= n:
end = mid - 1
else:
beg = mid + 1
return n - beg
``````
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