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This approach uses sorting to calculate the h-index. The idea is to sort the array of citations in descending order. Then, find the maximum number h such that there are h papers with at least h citations. This can be efficiently determined by iterating over the sorted array.
Time Complexity: O(n log n) due to sorting, Space Complexity: O(1) since the sorting is in place.
1#include <vector>
2#include <algorithm>
3#include <iostream>
4using namespace std;
5
6int hIndex(vector<int>& citations) {
7 sort(citations.begin(), citations.end(), greater<int>());
8 for (int i = 0; i < citations.size(); i++) {
9 if (citations[i] < i + 1) {
10 return i;
11 }
12 }
13 return citations.size();
14}
15
16int main() {
17 vector<int> citations = {3, 0, 6, 1, 5};
18 cout << "H-Index: " << hIndex(citations) << endl;
19 return 0;
20}This C++ implementation sorts the citations in descending order and finds the h-index using the same logic as the C solution.
Given the constraints where citation counts do not exceed 1000 and the number of papers is at most 5000, a counting sort or bucket sort can be used. This approach involves creating a frequency array to count citations. Then traverse the frequency array to compute the h-index efficiently.
Time Complexity: O(n + m) where n is citationsSize and m is the maximum citation value, Space Complexity: O(m).
1using System;
public class Solution {
public int HIndex(int[] citations) {
int n = citations.Length;
int[] count = new int[n + 1];
foreach (int c in citations) {
if (c >= n) count[n]++;
else count[c]++;
}
int total = 0;
for (int i = n; i >= 0; i--) {
total += count[i];
if (total >= i) return i;
}
return 0;
}
public static void Main(string[] args) {
int[] citations = {3, 0, 6, 1, 5};
Solution sol = new Solution();
Console.WriteLine("H-Index: " + sol.HIndex(citations));
}
}The C# solution aggregates citation frequencies within a fixed-sized array, iterating in reverse to determine the h-index.