Sum of Squares of Special Elements — Easy Problem & Solution

Problem statement

An element nums[i - 1] is special when i divides n, the length of nums. Note the 1-based index i.

Return the sum of the squares of all special elements.

Example 1

Input: nums = [1,2,3,4]
Output: 21
Explanation: n is 4, and 1, 2 and 4 divide it, giving 1² + 2² + 4² = 21.

Example 2

Input: nums = [2,7,1,19,18,3]
Output: 63
Explanation: n is 6, and 1, 2, 3 and 6 divide it, giving 4 + 49 + 1 + 9 = 63.

Example 3

Input: nums = [5]
Output: 25

Constraints

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

How to solve Sum of Squares of Special Elements

The condition is purely positional, so loop over 1-based positions, test divisibility against the length and square the value at that position.

Approach

1. Let n be the length of nums.
2. For i from 1 to n, check n % i == 0.
3. When it holds, add nums[i - 1] * nums[i - 1] to the total.

Why it works

Each position is tested exactly once against the exact predicate the statement names, so the sum ranges over precisely the special elements.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• Using the 0-based index in the divisibility test shifts every position by one and makes i = 0 a division by zero.
• Squaring values up to 50 across 50 positions stays well inside 32 bits, but squaring is easy to forget.

Reference solution

Python

from typing import List

def sumOfSquares(nums: List[int]) -> int:
    n = len(nums)
    return sum(nums[i - 1] ** 2 for i in range(1, n + 1) if n % i == 0)

JavaScript

var sumOfSquares = function(nums) {
    var n = nums.length, total = 0;
    for (var i = 1; i <= n; i++) {
        if (n % i === 0) total += nums[i - 1] * nums[i - 1];
    }
    return total;
};

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