Algorithm
Problem Name: 274. H-Index
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return compute the researcher's h-index.
According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.
If there are several possible values for h, the maximum one is taken as the h-index.
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.length1 <= n <= 50000 <= citations[i] <= 1000
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int hIndex(int* citations, int citationsSize) {
int i, j, k = 0;
int n;
#if 0 // O(n^2) 39ms
for (i = citationsSize; i > 0; i --) { // for every possible h
n = 0;
for (j = 0; j < citationsSize; j ++) { // for every citation
if (citations[j] >= i) n ++;
}
if (n >= i) {
return i;
}
}
#else // O(n) 3ms
int *x;
x = calloc(citationsSize + 1, sizeof(int));
//assert(x);
for (i = 0; i < citationsSize; i ++) { // for every citation
j = citations[i] > citationsSize ? citationsSize : citations[i];
x[j] ++;
}
n = 0;
for (i = citationsSize; i > 0; i --) {
n += x[i];
if (n >= i) { k = i; break; }
}
free(x);
#endif
return k;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int hIndex(vector<int>& citations) {
sort(citations.begin(), citations.end());
int i = 0, j = citations.size() - 1;
while(j >= 0 && citations[j] > i) i++, j--;
return i;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int hIndex(int[] citations) {
int n = citations.length;
Arrays.sort(citations);
for (int i = n - 1; i >= 0; i--) {
if (n - i > citations[i]) {
return n - i - 1;
}
}
return n;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const hIndex = function(citations) {
const n = citations.length
const arr = Array(n + 1).fill(0)
for(let e of citations) {
if(e >= n) arr[n]++
else arr[e]++
}
for(let i = n, sum = 0; i >= 0; i--) {
sum += arr[i]
if(sum >= i) return i
}
return 0
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def hIndex(self, citations):
citations.sort()
for i in range(len(citations)):
if len(citations) - i <= citations[i]: return len(citations) - i
return 0
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0274_HIndex
{
public int HIndex(int[] citations)
{
int n = citations.Length;
int[] papers = new int[n + 1];
foreach (int c in citations)
papers[Math.Min(n, c)]++;
int k = n;
for (int s = papers[n]; k > s; s += papers[k])
k--;
return k;
}
}
}
Input
Output