Sum of Squares of Special Elements

Problem Description

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

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

Solution Approach

Intuition:

To efficiently determine the sym of square of the special elements.

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

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

TypeScript

Written by @Ishitamukherjee2004

 class Solution {
sumOfSquares(nums: number[]): number {
 let n: number = nums.length;
 let sqr: number = 0;
 for (let i: number = 0; i < n; i++) {
   if (n % (i + 1) === 0) {
     sqr += nums[i] * nums[i];
   }
 }
 return sqr;
}
}

Python

Written by @Ishitamukherjee2004

 
class Solution:
def sumOfSquares(self, nums: List[int]) -> int:
 n = len(nums)
 sqr = 0
 for i in range(n):
   if n % (i + 1) == 0:
     sqr += nums[i] * nums[i]
 return sqr

Java

Written by @Ishitamukherjee2004

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

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 int sumOfSquares(vector<int>& nums) {
     int n = nums.size();
     int sqr = 0;
     for(int i=0; i<n; i++){
         if(n%(i+1)==0){
             sqr+=nums[i]*nums[i];
         }
     }
     return sqr;
 }
};

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$
• The time complexity is $O(n)$ where n is the length of the input array nums. This is because the function iterates through the array once, performing a constant amount of work for each element.
• The space complexity is $O(1)$ because we are not using any extra space.