H-index II

Time: O(LogN); Space: O(1); medium

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 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, 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?

Hints:

  1. Expected runtime complexity is in O(Log N) and the input is sorted.

1. Binary Search [O(LogN), O(1)]

[2]:

class Solution1(object):
    """
    Time: O(LogN)
    Space: O(1)
    """
    def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        n = len(citations)
        left, right = 0, n - 1

        while left <= right:
            mid = (left + right) // 2
            if citations[mid] >= n - mid:
                right = mid - 1
            else:
                left = mid + 1

        return n - left

[4]:

s = Solution1()

citations = [0,1,3,5,6]
assert s.hIndex(citations) == 3