H-Index
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, 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.
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
Approach #1 (Sorting)
Think geometrically. Imagine plotting a histogram where the y-axis represents the number of citations for each paper. After sorting in descending order, the h-index is the length of the largest square in the histogram.
Algorithm:
To find such a square length, we first sort the citations array in descending order. After sorting, if citations[i] > i, then papers 0 to i all have at least i+1 citations. Thus, to find the h-index, we search for the largest i (let's call it i') such that citations[i] > i and therefore the h-index is i'+1.
public class Solution {
public int hIndex(int[] citations) {
// sorting the citations in ascending order
Arrays.sort(citations);
// finding h-index by linear search
int i = 0;
while (i < citations.length && citations[citations.length - 1 - i] > i) {
i++;
}
return i; // after the while loop, i = i' + 1
}
}
Complexity Analysis
Time complexity : O(nlogn). Comparison sorting dominates the time complexity. Space complexity : O(1). Most libraries using heap sort which costs O(1) extra space in the worst case.
