# 0275. H-Index II

<https://leetcode.com/problems/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 an **ascending order**, return compute the researcher's `h`**-index**.

According to the [definition of h-index on Wikipedia](https://en.wikipedia.org/wiki/H-index): A scientist has an index `h` if `h` of their `n` papers have at least `h` citations each, and the other `n − h` papers have no more than `h` citations each.

If there are several possible values for `h`, the maximum one is taken as the `h`**-index**.

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**.

## ac

Binary search, please remember to follow standard template.

```java
class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        // edge case
        if (n == 0) return 0;

        int l = 0, r = n - 1;
        while (l + 1 < r) {
            int mid = l + (r-l)/2;
            if (citations[mid] >= n - mid) {
                r = mid;
            } else {
                l = mid;
            }
        }

        // compare left, right
        if (citations[l] >= n - l) {
            return n - l;
        } else if (citations[r] >= n - r) {
            return n - r;
        } else {
            return 0;
        }
    }
}
/**
N -> logN
Binary search
**/
```