275. H-Index II Medium

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 ascending order, return the researcher's h-index.

According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h papers that have each been cited at least h times.

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

Approach

Input: An array citations representing the number of citations for each of the researcher's papers.

Output: Return the h-index.

This problem belongs to the Binary Search on Boundary category.

We need to find the maximum h such that the researcher has at least h papers with >= h citations.

We can use binary search to narrow down the range. Let's assume h = len(citations) - mid.

• If citations[mid] >= len(citations) - mid, it means all papers from mid to the end have citation counts greater than or equal to len(citations) - mid. Therefore, the maximum h might be on the left side of mid (including mid), so we continue searching to the left by setting right = mid - 1.

• Conversely, if the citation count at mid is less than the remaining number of papers, it doesn't satisfy the definition of h. The maximum h must be on the right side of mid, so we continue searching to the right by setting left = mid + 1.

• Finally, the position where left ends up is the index corresponding to the maximum h. We can obtain the maximum h by len(citations) - left, which is the total number of papers with citations >= h.

• Suppose there is only one paper, then left == right == mid == 0. We still need to check if citations[mid] >= len(citations) - mid. Therefore, our terminating condition is left > right.

Implementation

python

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)  # Get the length of the citations array
        left, right = 0, n - 1  # Initialize the left and right pointers for binary search

        while left <= right:  # Continue the loop when the left pointer is less than or equal to the right pointer
            mid = (left + right) // 2  # Calculate the middle index

            # If the number of citations at the middle position >= the remaining number of papers (n - mid)
            # It means the h-index might be on the left side (including mid), move the right pointer to the left
            if citations[mid] >= n - mid:
                right = mid - 1
            # Otherwise, the h-index is on the right side, move the left pointer to the right
            else:
                left = mid + 1
        
        # Finally, left will stop at the boundary that satisfies the condition, n - left is the h-index
        # Because n - left means there are at least n - left papers with citations >= n - left
        return n - left

javascript

/**
 * @param {number[]} citations
 * @return {number}
 */
var hIndex = function(citations) {
    let left = 0;
    let right = citations.length - 1;

    while (left <= right) {
        const mid = Math.floor((left + right) / 2);

        if (citations[mid] >= citations.length - mid) {
            right = mid - 1;
        } else {
            left = mid + 1
        }
    }

    return citations.length - left;
};

Complexity Analysis

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

Links

275. H-Index II (English)

275. H 指数 II (Chinese)