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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.
1using System;
2
3public class Solution {
4 public int HIndex(int[] citations) {
5 Array.Sort(citations);
6 Array.Reverse(citations);
7 int n = citations.Length;
8 for (int i = 0; i < n; i++) {
9 if (citations[i] < i + 1) {
10 return i;
11 }
12 }
13 return n;
14 }
15
16 public static void Main(string[] args) {
17 int[] citations = {3, 0, 6, 1, 5};
18 Solution sol = new Solution();
19 Console.WriteLine("H-Index: " + sol.HIndex(citations));
20 }
21}This C# solution sorts the array in descending order and determines the h-index in a manner similar to the other solutions provided.
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).
1#include <vector>
#include <iostream>
using namespace std;
int hIndex(vector<int>& citations) {
int n = citations.size();
vector<int> count(n + 1, 0);
for (int c : 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;
}
int main() {
vector<int> citations = {3, 0, 6, 1, 5};
cout << "H-Index: " << hIndex(citations) << endl;
return 0;
}This C++ solution employs a vector to create a counting array, tracking citation frequencies.