Introduction

In this blog post, we will explore how to solve the LeetCode problem "274" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, 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. Example 1: Input: citations = [3,0,6,1,5] Output: 3 Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 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 = [1,3,1] Output: 1 Constraints: n == citations.length 1 <= n <= 5000 0 <= citations[i] <= 1000

Explanation

• Binary Search: The core idea is to use binary search to find the h-index. The search space is from 0 to n (the number of papers).
  • Check Condition: For each potential h-index value mid during the binary search, we check if there are at least mid papers with citations greater than or equal to mid.
  • Adjust Search Range: Based on the check, we adjust the binary search range (either the left or right boundary) to narrow down the search for the optimal h-index.

• Runtime Complexity: O(n log n), Storage Complexity: O(1)

Code

    def hIndex(citations):
    """
    Calculates the h-index of a researcher given an array of citations.

    Args:
        citations: A list of integers representing the number of citations for each paper.

    Returns:
        The h-index of the researcher.
    """
    n = len(citations)
    left, right = 0, n

    while left <= right:
        mid = (left + right) // 2
        count = 0
        for citation in citations:
            if citation >= mid:
                count += 1

        if count >= mid:
            left = mid + 1
        else:
            right = mid - 1

    return right