Welcome to Subscribe On Youtube

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;
    }
    
    

All Problems

All Solutions