LeetCode #274 — MEDIUM

H-Index

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, 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.

Example 1:

Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 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 = [1,3,1]
Output: 1

Constraints:

n == citations.length
1 <= n <= 5000
0 <= citations[i] <= 1000

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, 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.

Baseline thinking

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

Pattern signal: Array

Example 1

[3,0,6,1,5]

Example 2

[1,3,1]

Core Insight

What unlocks the optimal approach

An easy approach is to sort the array first.
What are the possible values of h-index?
A faster approach is to use extra space.

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 #274: H-Index
class Solution {
    public int hIndex(int[] citations) {
        Arrays.sort(citations);
        int n = citations.length;
        for (int h = n; h > 0; --h) {
            if (citations[n - h] >= h) {
                return h;
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #274: H-Index
func hIndex(citations []int) int {
	sort.Ints(citations)
	n := len(citations)
	for h := n; h > 0; h-- {
		if citations[n-h] >= h {
			return h
		}
	}
	return 0
}

# Accepted solution for LeetCode #274: H-Index
class Solution:
    def hIndex(self, citations: List[int]) -> int:
        citations.sort(reverse=True)
        for h in range(len(citations), 0, -1):
            if citations[h - 1] >= h:
                return h
        return 0

// Accepted solution for LeetCode #274: H-Index
impl Solution {
    #[allow(dead_code)]
    pub fn h_index(citations: Vec<i32>) -> i32 {
        let mut citations = citations;
        citations.sort_by(|&lhs, &rhs| rhs.cmp(&lhs));

        let n = citations.len();

        for i in (1..=n).rev() {
            if citations[i - 1] >= (i as i32) {
                return i as i32;
            }
        }

        0
    }
}

// Accepted solution for LeetCode #274: H-Index
function hIndex(citations: number[]): number {
    citations.sort((a, b) => b - a);
    for (let h = citations.length; h; --h) {
        if (citations[h - 1] >= h) {
            return h;
        }
    }
    return 0;
}

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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.