LeetCode #275 — MEDIUM

H-Index II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 non-descending 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.

Step 01

Brute Force Baseline

Problem summary: 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: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[0,1,3,5,6]

Example 2

[1,2,100]

Core Insight

What unlocks the optimal approach

Expected runtime complexity is in <i>O</i>(log <i>n</i>) and the input is sorted.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #275: H-Index II
class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (citations[mid] >= n - mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return n - left;
    }
}

// Accepted solution for LeetCode #275: H-Index II
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
}

# Accepted solution for LeetCode #275: H-Index II
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

// Accepted solution for LeetCode #275: H-Index II
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
    }
}

// Accepted solution for LeetCode #275: H-Index II
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.