275. H-Index II

1. Description

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.

2. Example

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

3. Constraints

• n == citations.length
• 1 <= n <= $10^5$
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

4. Solutions

Binary Search

n = citations.size()
Time complexity: O(logn)
Space complexity: O(1)

class Solution {
public:
    int hIndex(vector<int> &citations) {
        int result = 0;
        for (int left = 0, right = citations.size(); left < right;) {
            int mid = (left + right) / 2;
            if (citations[mid] >= right - mid &&
                (mid == 0 || citations[mid - 1] < right - (mid - 1))) {
                result = citations.size() - mid;
                break;
            } else if (citations[mid] >= citations.size() - mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return result;
    }
};

// shorter version

class Solution {
public:
    int hIndex(vector<int> &citations) {
        int left = 0, right = citations.size();
        for (; left < right;) {
            int mid = (left + right) / 2;
            if (citations[mid] >= citations.size() - mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return citations.size() - left;
    }
};