Introduction

In this blog post, we will explore how to solve the LeetCode problem "2778" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the problem:

• High-Level Approach:

  • Iterate from 1 to n (inclusive).
  • For each i, check if n is divisible by i.
  • If it is, add the square of nums[i-1] to the running sum.

• Complexity Analysis:

  • Runtime Complexity: O(n) - We iterate through the array once.
  • Storage Complexity: O(1) - We only use a constant amount of extra space.

Code

    def sum_of_special_elements(nums):
    """
    Calculates the sum of the squares of special elements in the input array.

    Args:
        nums: A 1-indexed integer array.

    Returns:
        The sum of the squares of all special elements.
    """
    n = len(nums)
    total_sum = 0
    for i in range(1, n + 1):
        if n % i == 0:
            total_sum += nums[i - 1] * nums[i - 1]
    return total_sum