Question

275. H-Index II

中文文档

Description

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 ascending order, 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.

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.

Solutions

Solution 1: Binary Search

We notice that if there are at least $x$ papers with citation counts greater than or equal to $x$, then for any $y \lt x$, its citation count must also be greater than or equal to $y$. This exhibits monotonicity.

Therefore, we use binary search to enumerate $h$ and obtain the maximum $h$ that satisfies the condition. Since we need to satisfy that $h$ papers are cited at least $h$ times, we have $citations[n - mid] \ge mid$.

The time complexity is $O(\log n)$, where $n$ is the length of the array $citations$. The space complexity is $O(1)$.

class Solution:
  def hIndex(self, citations: List[int]) -> int:
    n = len(citations)
    left, right = 0, n
    while left < right:
      mid = (left + right + 1) >> 1
      if citations[n - mid] >= mid:
        left = mid
      else:
        right = mid - 1
    return left

class Solution {
  public int hIndex(int[] citations) {
    int n = citations.length;
    int left = 0, right = n;
    while (left < right) {
      int mid = (left + right) >>> 1;
      if (citations[mid] >= n - mid) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return n - left;
  }
}

class Solution {
public:
  int hIndex(vector<int>& citations) {
    int n = citations.size();
    int left = 0, right = n;
    while (left < right) {
      int mid = (left + right + 1) >> 1;
      if (citations[n - mid] >= mid)
        left = mid;
      else
        right = mid - 1;
    }
    return left;
  }
};

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

function hIndex(citations: number[]): number {
  const n = citations.length;
  let left = 0,
    right = n;
  while (left < right) {
    const mid = (left + right + 1) >> 1;
    if (citations[n - mid] >= mid) {
      left = mid;
    } else {
      right = mid - 1;
    }
  }
  return left;
}

impl Solution {
  pub fn h_index(citations: Vec<i32>) -> i32 {
    let n = citations.len();
    let (mut left, mut right) = (0, n);
    while left < right {
      let mid = ((left + right + 1) >> 1) as usize;
      if citations[n - mid] >= (mid as i32) {
        left = mid;
      } else {
        right = mid - 1;
      }
    }
    left as i32
  }
}

public class Solution {
  public int HIndex(int[] citations) {
    int n = citations.Length;
    int left = 0, right = n;
    while (left < right) {
      int mid = (left + right + 1) >> 1;
      if (citations[n - mid] >= mid) {
        left = mid;
      } else {
        right = mid - 1;
      }
    }
    return left;
  }
}