# H-Index II
**Difficulty:** MEDIUM
[External](https://leetcode.com/problems/h-index-ii)
Canonical: https://scaleengineer.com/dsa/problems/h-index-ii
**Algorithms:** [Binary Search](https://scaleengineer.com/algorithms/binary-search)
**Data structures:** Array
---
## Problem
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 **non-descending order**, return _the researcher's h-index_.

According to the [definition of h-index on Wikipedia](https://en.wikipedia.org/wiki/H-index): 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**.

# Approaches
## Linear Search Approach
Iterate through the array and for each index check if it satisfies the h-index condition.
**Time:** O(n) where n is the length of the array · **Space:** O(1) constant space
**Pros:** Simple to understand and implement; Works well for small arrays; No extra space required
**Cons:** Does not utilize the sorted nature of the array; Not optimal for large arrays; Does not meet the requirement of logarithmic time complexity
### Explanation
This approach involves iterating through the array from left to right. For each position i, we check if citations[i] is greater than or equal to the number of papers remaining (n-i). The first position where this condition is true gives us our h-index.

```java
class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        for (int i = 0; i < n; i++) {
            int h = n - i;
            if (citations[i] >= h) {
                return h;
            }
        }
        return 0;
    }
}
```

The idea is that at each index i, we have n-i papers remaining (including the current one). If the current citation count is greater than or equal to the remaining papers, we've found our h-index.
### Algorithm
1. Get the length of the array n
2. Iterate through the array from index 0 to n-1
3. For each index i:
   - Calculate h = n - i (remaining papers)
   - If citations[i] >= h, return h
4. If no h-index found, return 0

## Binary Search Approach
Use binary search to find the h-index by exploiting the sorted nature of the array.
**Time:** O(log n) where n is the length of the array · **Space:** O(1) constant space
**Pros:** Meets the logarithmic time requirement; Efficiently utilizes the sorted nature of the array; Works well for large arrays
**Cons:** Slightly more complex to implement than linear search; Binary search implementation needs to be careful to handle edge cases
### Explanation
Since the array is sorted in ascending order, we can use binary search to find the h-index. The key insight is that for any index i, if citations[i] >= n-i, then n-i is a potential h-index. We want to find the largest such value.

```java
class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int left = 0;
        int right = n - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            int h = n - mid;
            
            if (citations[mid] == h) {
                return h;
            } else if (citations[mid] < h) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return n - left;
    }
}
```

The binary search maintains a window where the h-index could be. At each step:
- Calculate the middle point
- Compare citations[mid] with n-mid (potential h-index)
- If citations[mid] is too small, search right half
- If citations[mid] is too large, search left half
### Algorithm
1. Initialize left = 0 and right = n-1
2. While left <= right:
   - Calculate mid = left + (right - left) / 2
   - Calculate potential h-index h = n - mid
   - If citations[mid] == h, return h
   - If citations[mid] < h, search right half (left = mid + 1)
   - If citations[mid] > h, search left half (right = mid - 1)
3. Return n - left

# Solutions
### Java

```java
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;
  }
}

```

### CSharp

```csharp
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;
    }
}
```

### CPP

```cpp
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;
  }
};

```

### Python

```python
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
```