LeetCode 2778. Sum of Squares of Special Elements Solution in Java, C++, Python & More | Explanation + Code

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 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

Solutions

Solution 1

PythonJavaC++GoTypeScript

class Solution: def sumOfSquares(self, nums: List[int]) -> int: n = len(nums) return sum(x * x for i, x in enumerate(nums, 1) if n % i == 0)(code-box)

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

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

func sumOfSquares(nums []int) (ans int) { n := len(nums) for i, x := range nums { if n%(i+1) == 0 { ans += x * x } } return }(code-box)

function sumOfSquares(nums: number[]): number { const n = nums.length; let ans = 0; for (let i = 0; i < n; ++i) { if (n % (i + 1) === 0) { ans += nums[i] * nums[i]; } } return ans; }(code-box)