H-Index LeetCode Solution | Easy Approach

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H-Index Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return compute the researcher’s h-index.

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

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

H-Index Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  int hIndex(vector<int>& citations) {
    const int n = citations.size();
    int accumulate = 0;
    vector<int> count(n + 1);

    for (const int citation : citations)
      ++count[min(citation, n)];

    // to find the largeset h-index, loop from back to front
    // i is the candidate h-index
    for (int i = n; i >= 0; --i) {
      accumulate += count[i];
      if (accumulate >= i)
        return i;
    }

    throw;
  }
};

Java

 class Solution {
  public int hIndex(int[] citations) {
    final int n = citations.length;
    int accumulate = 0;
    int[] count = new int[n + 1];

    for (final int citation : citations)
      ++count[Math.min(citation, n)];

    // to find the largeset h-index, loop from back to front
    // i is the candidate h-index
    for (int i = n; i >= 0; --i) {
      accumulate += count[i];
      if (accumulate >= i)
        return i;
    }

    throw new IllegalArgumentException();
  }
}

Python

 class Solution:
  def hIndex(self, citations: List[int]) -> int:
    n = len(citations)
    accumulate = 0
    count = [0] * (n + 1)

    for citation in citations:
      count[min(citation, n)] += 1

    # to find the largeset h-index, loop from back to front
    # i is the candidate h-index
    for i, c in reversed(list(enumerate(count))):
      accumulate += c
      if accumulate >= i:
        return i

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