Sum of Squares of Special Elements

Problem

You are given a 1-indexed integer array nums of length n.

An element nums[i] of nums is called special if i divides n, i.e. n % i == 0.

Return **the *sum of the squares* of all special elements of **nums.

Example 1:

Input: nums = [1,2,3,4]
Output: 21
Explanation: There are exactly 3 special elements in nums: nums[1] since 1 divides 4, nums[2] since 2 divides 4, and nums[4] since 4 divides 4.
Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21.

Example 2:

Input: nums = [2,7,1,19,18,3]
Output: 63
Explanation: There are exactly 4 special elements in nums: nums[1] since 1 divides 6, nums[2] since 2 divides 6, nums[3] since 3 divides 6, and nums[6] since 6 divides 6.
Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63.

Constraints:

1 <= nums.length == n <= 50
1 <= nums[i] <= 50

Solution (Java)

class Solution {
    public int sumOfSquares(int[] nums) {
        int sum = 0;
        for(int i = 0; i < nums.length; i ++) {
            if(nums.length % (i + 1) == 0)
            sum += nums[i] * nums[i];
        }
        return sum;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).