Time Complexity: O(N), where N is the number being checked. This is because the loop runs from 1 to N to find all the divisors.

Space Complexity: O(1), as no extra space is used except for a few integer variables. The check is done in constant space.

Optimal Approach

Algorithm

Traverse from 1 to square root of n.
Maintain a variable sum to calculate sum of divisors.
If i is a factor of n,add it to the sum as well as add n/i to the sum.But don’t forget to check if i and n/i are same or not.For instance if i=6 and n=36,then i and n/i will have same value and it will not make sense to add both i and n/i,i.e 6 two times.

Code

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