# 275. H-Index II

## 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 {
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[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;
}


• 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
}
}