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275. HIndex II
Description
Given an array of integers citations
where citations[i]
is the number of citations a researcher received for their i^{th}
paper and citations
is sorted in ascending order, return the researcher's hindex.
According to the definition of hindex on Wikipedia: The hindex 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 hindex is 3.
Example 2:
Input: citations = [1,2,100] Output: 2
Constraints:
n == citations.length
1 <= n <= 10^{5}
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 { 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; } }

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; } };

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

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

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; } }

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 } }