Algorithm
Problem Name: 275. H-Index II
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in an ascending order, 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.
You must write an algorithm that runs in logarithmic time.
Example 1:
Input: citations = [0,1,3,5,6] Output: 3 Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had received 0, 1, 3, 5, 6 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 = [1,2,100] Output: 2
Constraints:
n == citations.length1 <= n <= 1050 <= citations[i] <= 1000citationsis sorted in ascending order.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int hIndex(int* citations, int citationsSize) {
int i, h = 0;
for (i = 0; i < citationsSize; i ++) {
if (citations[citationsSize - 1 - i] >= i + 1) {
h = i + 1;
} else {
break;
}
}
return h;
}
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Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int hIndex(vector<int>& citations) {
int i = 0, j = citations.size() - 1;
while(j >= 0 && citations[j] > i) i++, j--;
return i;
}
};
// Binary Search, O(logn)
class Solution {
public:
int hIndex(vector<int>& citations) {
int lo = 0, len = citations.size(), hi = len - 1;
int mid = lo + (hi - lo) / 2;
while(lo < = hi){
if(citations[mid] >= len - mid) hi = mid - 1;
else lo = mid + 1;
mid = lo + (hi - lo) / 2;
}
return len - lo;
}
};
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Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int hIndex(int[] citations) {
if (citations.length == 0) {
return 0;
}
if (citations.length == 1) {
return citations[0] > 0 ? 1 : 0;
}
return binarySearch(citations, 0, citations.length - 1);
}
private int binarySearch(int[] citations, int start, int end) {
int ind = -1;
int n = end + 1;
while (start < end) {
int mid = (start + end) / 2 + 1;
if (citations[mid] > n - mid) {
end = mid - 1;
}
else {
start = mid;
}
}
if (citations[start] > n - start) {
return n;
}
return citations[start] == n - start ? n - start : n - start - 1;
}
}
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Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const hIndex = function(citations) {
let left = 0,
len = citations.length,
right = len - 1,
mid
while (left <= right) {
mid = left + ((right - left) >> 1)
if (citations[mid] >= len - mid) right = mid - 1
else left = mid + 1
}
return len - left
}
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Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def hIndex(self, citations):
l, r, res = 0, len(citations) - 1, 0
while l <= r:
mid = (l + r) // 2
if len(citations) - mid <= citations[mid]: res, r = len(citations) - mid, r - 1
else: l = mid + 1
return res
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Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0275_HIndexII
{
public int HIndex(int[] citations)
{
int n = citations.Length;
int left = 0, right = n - 1;
while (left < = right)
{
var mid = left + (right - left) / 2;
if (citations[mid] == n - mid) return n - mid;
else if (citations[mid] < n - mid) left = mid + 1;
else right = mid - 1;
}
return n - left;
}
}
}
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Output