H-Index II

Given an array of citations sorted in ascending order (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.

According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."

Example:

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, 
             her h-index is 3.

Note:

If there are several possible values for h, the maximum one is taken as the h-index.

 

class Solution {
    public int hIndex(int[] citations) {
        int length = citations.length;

    int low = 0, high = length - 1, mid;

    while (low <= high) {
        mid = (low + high) / 2;

        if (length - mid > citations[mid])
            low = mid + 1;
        else
            high = mid - 1;
    }

    return length - low;
    }
}

Try it on Leetcode

Formally, if f is the function that corresponds to the number of citations for each publication, we compute the h-index as follows: First we order the values of f from the largest to the lowest value. Then, we look for the last position in which f is greater than or equal to the position (we call h this position). For example, if we have a researcher with 5 publications A, B, C, D, and E with 10, 8, 5, 4, and 3 citations, respectively, the h-index is equal to 4 because the 4th publication has 4 citations and the 5th has only 3. In contrast, if the same publications have 25, 8, 5, 3, and 3 citations, then the index is 3 because the fourth paper has only 3 citations.

f(A)=10, f(B)=8, f(C)=5, f(D)=4, f(E)=3　→ h-index=4
f(A)=25, f(B)=8, f(C)=5, f(D)=3, f(E)=3　→ h-index=3
[Above explanation is taken from H-index]

From above explanation, it is easy to understand that,[citation sorted in descending order] if index position is greater than or equal to citation then particular position is H-index. We can easily find by iterating each element but it takes O(N). But we solved using O(log N).

We are going to find a position by subtracting low or mid from total length.

1) If length-mid is greater than citation then, increase mid by 1 and store it in low.
2) Else, decrease mid by 1 and store it in high
3) Repeat above steps until low is greater than high.
4) Finally return length-low, which will be h-index.

Time Complexity: O(log N)

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