# Question

Formatted question description: https://leetcode.ca/all/275.html

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 ascending order, return the researcher's h-index.

According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h papers that have each been cited at least h times.

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.length
• 1 <= n <= 105
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

# Algorithm

First initialize left and right to 0 and len-1, then take the middle value mid, compare citations[mid] with len-mid,

• If the former is big, right moves before mid
• On the contrary, right moves after mid The loop condition is left<=right, and finally returns len-left.

# Code

• 
public class H_Index_II {
class Solution {
public int hIndex(int[] citations) {
int len = citations.length, left = 0, right = len - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (citations[mid] == len - mid) {
return len - mid;
}
else if (citations[mid] > len - mid) { // cites > h, try next round
right = mid - 1;
}
else {
left = mid + 1;
}
}

return len - left;
}
}
}

############

class Solution {
public int hIndex(int[] citations) {
int n = citations.length;
int left = 0, right = n;
while (left < right) {
int mid = (left + right + 1) >> 1;
if (citations[n - mid] >= mid) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}
}

• // OJ: https://leetcode.com/problems/h-index-ii/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
int hIndex(vector<int>& A) {
int N = A.size(), L = 0, R = N - 1;
while (L <= R) {
int M = (L + R) / 2;
if (A[M] >= N - M) R = M - 1;
else L = M + 1;
}
return N - L;
}
};

• class Solution:
def hIndex(self, citations: List[int]) -> int:
n = len(citations)
left, right = 0, n
while left < right:
mid = (left + right + 1) >> 1
if citations[n - mid] >= mid:
left = mid
else:
right = mid - 1
return left

############

class Solution(object):
def hIndex(self, citations):
"""
:type citations: List[int]
:rtype: int
"""
if not citations:
return 0
n = len(citations)
start, end = 0, n - 1
while start < end:
mid = start + (end - start) / 2
if citations[mid] >= n - mid:
end = mid
else:
start = mid + 1
return n - start if citations[start] != 0 else 0


• func hIndex(citations []int) int {
n := len(citations)
left, right := 0, n
for left < right {
mid := (left + right + 1) >> 1
if citations[n-mid] >= mid {
left = mid
} else {
right = mid - 1
}
}
return left
}

• function hIndex(citations: number[]): number {
const n = citations.length;
let left = 0,
right = n;
while (left < right) {
const mid = (left + right + 1) >> 1;
if (citations[n - mid] >= mid) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

`