Compute the H-Index From Citation Counts
Problem
Given citation counts for a researcher's papers, return the h-index — the largest h such that at least h papers have been cited at least h times each. The clean linear-time approach: bucket each paper by its citation count, capping at n (any count above n is the same as n since you can't have more than n papers). Then walk buckets from n down to 0, accumulating the running tally of papers with at least that many citations. The first bucket where that tally is at least the bucket index is the h-index.
Input
citations = [3, 0, 6, 1, 5]Output
33 papers have ≥ 3 citations (the ones with 3, 6, 5). No 4 papers have ≥ 4 citations. So h = 3.
def h_index(citations):
n = len(citations)
buckets = [0] * (n + 1)
for c in citations:
buckets[min(c, n)] += 1
count = 0
for h in range(n, -1, -1):
count += buckets[h]
if count >= h:
return h
return 0function hIndex(citations) {
const n = citations.length;
const buckets = new Array(n + 1).fill(0);
for (const c of citations) buckets[Math.min(c, n)]++;
let count = 0;
for (let h = n; h >= 0; h--) {
count += buckets[h];
if (count >= h) return h;
}
return 0;
}class Solution {
public int hIndex(int[] citations) {
int n = citations.length;
int[] buckets = new int[n + 1];
for (int c : citations) buckets[Math.min(c, n)]++;
int count = 0;
for (int h = n; h >= 0; h--) {
count += buckets[h];
if (count >= h) return h;
}
return 0;
}
}int hIndex(vector<int>& citations) {
int n = citations.size();
vector<int> buckets(n + 1, 0);
for (int c : citations) buckets[min(c, n)]++;
int count = 0;
for (int h = n; h >= 0; h--) {
count += buckets[h];
if (count >= h) return h;
}
return 0;
}