H-Index II - Practice Coding Problems

H-Index II - Problem

The H-Index is a fascinating metric used to measure a researcher's academic impact! 🎓

Given a sorted array of integers citations where citations[i] represents the number of citations a researcher received for their i-th paper, your task is to find their H-Index.

What is H-Index?
The H-Index is defined as the maximum value of h such that the researcher has published at least h papers that have each been cited at least h times.

The Challenge: Since the array is already sorted, you must write an algorithm that runs in O(log n) time - think binary search! 🔍

Example: If a researcher has papers with citations [0, 1, 3, 5, 6], they have 3 papers with at least 3 citations each (papers with 3, 5, and 6 citations), so their H-Index is 3.

Input & Output

example_1.py — Basic Case

$ Input: citations = [0,1,3,5,6]

› Output: 3

💡 Note: The researcher has 3 papers with at least 3 citations each (papers with 3, 5, and 6 citations), and the remaining 2 papers have ≤ 3 citations. So H-Index = 3.

example_2.py — All High Citations

$ Input: citations = [1,2,100]

› Output: 2

💡 Note: The researcher has 2 papers with at least 2 citations each (papers with 2 and 100 citations). Can't have H-Index = 3 because there are only 3 papers total, but only 1 paper has ≥ 3 citations.

example_3.py — All Zero Citations

$ Input: citations = [0,0,0,0]

› Output: 0

💡 Note: No papers have any citations, so the H-Index is 0. The researcher has 0 papers with at least 1 citation each.

Constraints

1 ≤ citations.length ≤ 10⁵
0 ≤ citations[i] ≤ 1000
citations is sorted in non-decreasing order
Follow up: This is a follow-up problem to H-Index, where citations is not sorted.

Visualization

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Understanding the Visualization

Understanding the Problem

We need to find the maximum h where we have ≥h papers with ≥h citations each

Key Insight

In a sorted array, position i has exactly (n-i) papers from i to the end

Binary Search Logic

If citations[i] ≥ (n-i), then H-Index could be (n-i) or larger

Final Result

Binary search finds the optimal position efficiently

Key Takeaway

🎯 Key Insight: Binary search finds the optimal H-Index by leveraging the sorted array property to locate the transition point efficiently in O(log n) time.

Asked in

G Google 45 f Facebook 32 a Amazon 28 ⊞ Microsoft 24

The optimal solution uses binary search to achieve O(log n) time complexity. The key insight is that for any position i in the sorted array, we have exactly (n-i) papers from that position to the end. If citations[i] >= (n-i), then we have a valid H-Index of (n-i). Binary search helps us find the maximum such valid H-Index efficiently.

Common Approaches

Approach	Time	Space	Notes
✓ Binary Search (Optimal)	O(log n)	O(1)	Use binary search to find the maximum valid H-Index in O(log n) time
Brute Force (Linear Search)	O(n²)	O(1)	Check every possible H-Index value from 1 to n

Binary Search (Optimal) — Algorithm Steps

Initialize left=0, right=n-1 for binary search bounds
For each middle position, calculate papers_count = n - mid
If citations[mid] >= papers_count, we found a valid H-Index, search left half for potentially larger one
Otherwise, search right half for smaller valid H-Index
Return the maximum valid H-Index found

Visualization

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Step-by-Step Walkthrough

Initial Setup

Array [0,1,3,5,6], left=0, right=4

Check Mid=2

citations[2]=3, papers_count=3, 3≥3 ✓, search left half

Check Mid=0

citations[0]=0, papers_count=5, 0<5 ✗, search right half

Found Answer

left=2, H-Index = n-left = 5-2 = 3

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int hIndex(int* citations, int citationsSize) {
    int left = 0, right = citationsSize - 1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        int papersCount = citationsSize - mid; // Number of papers from mid to end
        
        if (citations[mid] >= papersCount) {
            // Found valid H-Index, check if we can find larger one
            right = mid - 1;
        } else {
            // Need more citations, move to right half
            left = mid + 1;
        }
    }
    
    return citationsSize - left; // Convert position back to H-Index
}

int main() {
    int citations[] = {0, 1, 3, 5, 6};
    int size = 5;
    printf("%d\n", hIndex(citations, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Binary search reduces search space by half each iteration

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for binary search pointers

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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