2728. Sum of Squares of Special Elements

2728. Sum of Squares of Special Elements

Difficulty: Easy

Topics: Mid Level, Array, Enumeration, Weekly Contest 354

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.

Example 3:

• Input: nums = [5]
• Output: 25

Example 4:

• Input: nums = [1,2]
• Output: 5

Example 5:

• Input: nums = [50,50]
• Output: 5000

Constraints:

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

Hint:

1. Iterate over all the elements of the array. For each index i, check if it is special using the modulo operator.
2. if n%i == 0, index i is special, and you should add nums[i] to the answer.

Similar Questions:

1. 633. Sum of Square Numbers
2. 1588. Sum of All Odd Length Subarrays

Solution:

We solve the problem by iterating over the 1-indexed positions of nums, checking whether each position i divides the array length n. If it does, we add the square of the corresponding element to the total sum.

Approach

• Let n = count($nums).
• Iterate i from 1 to n, because the problem uses 1-based indexing.
• Check if n % i == 0.
• If true, i is a special index.
• Convert the 1-based index to PHP’s 0-based index using $i - 1.
• Add nums[$i - 1] * nums[$i - 1] to the result.
• Return the accumulated sum.

Let's implement this solution in PHP: 2728. Sum of Squares of Special Elements

<?php
/**
 * @param Integer[] $nums
 * @return Integer
 */
function sumOfSquares(array $nums): int
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo sumOfSquares([1,2,3,4]) .  "\n";               // Output: 21
echo sumOfSquares([2,7,1,19,18,3]) .  "\n";         // Output: 63
echo sumOfSquares([5]) .  "\n";                     // Output: 25
echo sumOfSquares([1,2]) .  "\n";                   // Output: 5
echo sumOfSquares([50,50]) .  "\n";                 // Output: 5000
?>

Explanation:

• In PHP, arrays start at index 0, but the problem defines special positions using 1-based indexing.
• For example, nums[1] in the problem corresponds to $nums[0] in PHP.
• The condition n % i == 0 checks whether position i divides the total length n.
• For each valid special position, we square the value and add it to the answer.
• This directly follows the hint: check divisibility and accumulate squared values.

Complexity Analysis

• Time Complexity: O(n) — we check each index from 1 to n exactly once.
• Space Complexity: O(1) — only a few scalar variables are used.

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