H-Index II LeetCode Solution | Easy Approach

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H-Index II 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 an ascending order, 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.

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

Constraints:

  • n == citations.length
  • 1 <= n <= 105
  • 0 <= citations[i] <= 1000
  • citations is sorted in ascending order.

H-Index II Solutions

Time: O(log n)
Space: O(1)

C++

class Solution {
 public:
  int hIndex(vector<int>& citations) {
    int l = 0;
    int r = citations.size();

    while (l < r) {
      const int m = l + (r - l) / 2;
      if (citations[m] >= citations.size() - m)
        r = m;
      else
        l = m + 1;
    }

    return citations.size() - l;
  }
};

Java

 
class Solution {
  public int hIndex(int[] citations) {
    int l = 0;
    int r = citations.length;

    while (l < r) {
      final int m = l + (r - l) / 2;
      if (citations[m] >= citations.length - m)
        r = m;
      else
        l = m + 1;
    }

    return citations.length - l;
  }
}

Python

  class Solution:
  def hIndex(self, citations: List[int]) -> int:
    l = 0
    r = len(citations)

    while l < r:
      m = (l + r) // 2
      if citations[m] >= len(citations) - m:
        r = m
      else:
        l = m + 1

    return len(citations) - l

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