Problem
Given an array of integers citations
where citations[i]
is the number of citations a researcher received for their ith
paper, 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.
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.length
1 <= n <= 5000
0 <= citations[i] <= 1000
Solution
class Solution {
public int hIndex(int[] citations) {
// Sort array then traverse from end keep track of counter denoting total elements seen so
// far
Arrays.sort(citations);
int count = 0;
int hIndex = 0;
for (int i = citations.length - 1; i >= 0; i--) {
if (i == citations.length - 1 && count == citations[i]) {
hIndex = citations[i];
return hIndex;
// Ex:- 7 10--> counter =8
} else if (citations[i] <= count && count < citations[i + 1]) {
hIndex = count;
return hIndex;
// Ex:- 7 9 --> counter 6 (incuding 7 there willbe 7 elements)
} else if (citations[i] == count + 1) {
hIndex = count + 1;
return hIndex;
} else {
count++;
}
}
// case when no element is hindex so far
if (count < citations[0]) {
hIndex = count;
}
return hIndex;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).