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Question

Formatted question description: https://leetcode.ca/all/274.html

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

Algorithm

  1. Sort all SCI papers published by him in descending order of the number of citations;
  2. Search the sorted list from front to back until the serial number of a paper is greater than the number of times the paper has been cited.

The obtained serial number minus one is the H index

Code

  • 
public class H_Index {
    class Solution {
        public int hIndex(int[] citations) {

            if (citations == null || citations.length == 0) {
                return 0;
            }

            // @note
            Arrays.sort(citations); // 6,5,3,1,0
            for (int i = 0; i < citations.length / 2; ++i) {
                int temp = citations[i];
                citations[i] = citations[citations.length - i - 1];
                citations[citations.length - i - 1] = temp;
            }

            for (int i = 0; i < citations.length; ++i) {
                if (citations[i] <= i) { // @note: <i then return i+1 will not work
                    return i;
                }
            }

            return citations.length;

        }
    }
}


############

class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int[] cnt = new int[n + 1];
        for (int c : citations) {
            if (c <= n) {
                ++cnt[c];
            } else {
                ++cnt[n];
            }
        }
        int sum = 0;
        for (int i = n; i >= 0; --i) {
            sum += cnt[i];
            if (sum >= i) {
                return i;
            }
        }
        return 0;
    }
}


  • // OJ: https://leetcode.com/problems/h-index/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int hIndex(vector<int>& A) {
        sort(begin(A), end(A));
        int L = 0, R = A.size() - 1, N = A.size();
        while (L <= R) {
            int M = (L + R) / 2;
            if (A[M] < N - M) L = M + 1;
            else R = M - 1;
        }
        return N - L;
    }
};

  • class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)
        cnt = [0] * (n + 1)
        for c in citations:
            if c <= n:
                cnt[c] += 1
            else:
                cnt[n] += 1
        sum = 0
        for i in range(n, -1, -1):
            sum += cnt[i]
            if sum >= i:
                return i
        return 0

############

class Solution(object):
  def hIndex(self, citations):
    """
    :type citations: List[int]
    :rtype: int
    """
    n = len(citations)
    dp = [0] * (n + 1)
    for c in citations:
      if c > n:
        dp[n] += 1
      else:
        dp[c] += 1

    total = 0
    for i in reversed(range(1, len(dp))):
      total += dp[i]
      if total >= i:
        return i
    return 0


  • func hIndex(citations []int) int {
	n := len(citations)
	left, right := 0, n
	for left+1 < right {
		mid := int(uint(left+right) >> 1)
		if check(citations, mid) {
			left = mid
		} else {
			right = mid
		}
	}
	if check(citations, right) {
		return right
	}
	return left
}

func check(citations []int, mid int) bool {
	cnt := 0
	for _, citation := range citations {
		if citation >= mid {
			cnt++
		}
	}
	return cnt >= mid
}


  • function hIndex(citations: number[]): number {
    let n = citations.length;
    let cnt = new Array(n + 1).fill(0);
    for (let c of citations) {
        if (c <= n) {
            ++cnt[c];
        } else {
            ++cnt[n];
        }
    }
    let sum = 0;
    for (let i = n; i > -1; --i) {
        sum += cnt[i];
        if (sum >= i) {
            return i;
        }
    }
    return 0;
}

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