Problem URL: H-index II

We will seach for the highest mid where the value of mid is greater that or equal to the numbers of elements on the right from that mid index.

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        if citations[-1] == 0:
            return 0
        n = len(citations) 
        l, r = 0, n-1
        while l < r:
            m = (l+r)//2
            if citations[m] >= n-m:
                r = m
            else:
                l = m+1
        return n-l

Time Complexity: O(log(n))
Space Complexity: O(1)