Question

Formatted question description: https://leetcode.ca/all/275.html

 275	H-Index II

 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:

 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?

Algorithm

First initialize left and right to 0 and len-1, then take the middle value mid, compare citations[mid] with len-mid,

  • If the former is big, right moves before mid
  • On the contrary, right moves after mid The loop condition is left<=right, and finally returns len-left.

Code

Java

  • 
    public class H_Index_II {
        class Solution {
            public int hIndex(int[] citations) {
                int len = citations.length, left = 0, right = len - 1;
                while (left <= right) {
                    int mid = (left + right) / 2;
                    if (citations[mid] == len - mid) {
                        return len - mid;
                    }
                    else if (citations[mid] > len - mid) { // cites > h`, try next round
                        right = mid - 1;
                    }
                    else {
                        left = mid + 1;
                    }
                }
    
                return len - left;
            }
        }
    }
    
    
  • // OJ: https://leetcode.com/problems/h-index-ii/
    // Time: O(logN)
    // Space: O(1)
    class Solution {
    public:
        int hIndex(vector<int>& A) {
            int N = A.size(), L = 0, R = N - 1;
            while (L <= R) {
                int M = (L + R) / 2;
                if (A[M] >= N - M) R = M - 1;
                else L = M + 1;
            }
            return N - L;
        }
    };
    
  • class Solution(object):
      def hIndex(self, citations):
        """
        :type citations: List[int]
        :rtype: int
        """
        if not citations:
          return 0
        n = len(citations)
        start, end = 0, n - 1
        while start < end:
          mid = start + (end - start) / 2
          if citations[mid] >= n - mid:
            end = mid
          else:
            start = mid + 1
        return n - start if citations[start] != 0 else 0
    
    

All Problems

All Solutions