Sum of Squares at Divisor Indices
IntuitionThe problem requires identifying indices (1-based) that divide the length of the list evenly. These indices are "special positions". We then square the
# Intuition
The problem requires identifying indices (1-based) that divide the length of the list evenly. These indices are "special positions". We then square the elements at these positions and add them up. The task is essentially about looping through valid divisors and summing up squares.
# Approach
- Compute the length l of the list.
- Iterate through i from 1 to l:
- If l % i == 0, it's a valid divisor.
- Add the square of n[i - 1] to the sum (i-1 because of 0-based indexing).
- Return the final sum.

# Complexity
- Time complexity:
O(n)
You loop from 1 to n to check for divisors, which takes linear time in the size of the list.

- Space complexity:
O(1)
Only a constant amount of space is used (just variables s, l, and loop counter i).

# Code
```python []
class Solution(object):
    def sumOfSquares(self, n):
        l=len(n)
        s=0
        for i in range(1,l+1):
            if l%i==0:
                s+=n[i-1]**2
        return s
```
['Python']