Given an array of citations sorted in ascending order (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 = [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, her h-index is 3.
Note:
If there are several possible values for h, the maximum one is taken as the h-index.
Follow up:
       This is a follow up problem to H-Index, where citations is now guaranteed to be sorted in ascending order.
       Could you solve it in logarithmic time complexity?
class Solution:
     """
     time: O(logn)
     """
     def hIndex(self, citations):
         """
         :type citations: List[int]
         :rtype: int
         """
         if not citations:
             return 0
         
         n = len(citations)
         left, right = 0, n - 1
         
         while left + 1 < right:
             mid = (left + right) // 2
             
             if citations[mid] >= n - mid:
                 right = mid
             else:
                 left = mid
         
         for mid in (left, right):
             if citations[mid] >= n - mid:
                 return n - mid
         
         return 0
class Solution:
     """
     time: O(logn)
     """
     def hIndex(self, citations):
         """
         :type citations: List[int]
         :rtype: int
         """
         if not citations:
             return 0
         
         n = len(citations)
         left, right = 0, n - 1
         
         while left + 1 < right:
             mid = (left + right) // 2
             
             if citations[mid] >= n - mid:
                 right = mid
             else:
                 left = mid
         
         for mid in (left, right):
             if citations[mid] >= n - mid:
                 return n - mid
         
         return 0