Introduction

In this blog post, we will explore how to solve the LeetCode problem "1390" 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

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors. If there is no such integer in the array, return 0. Example 1: Input: nums = [21,4,7] Output: 32 Explanation: 21 has 4 divisors: 1, 3, 7, 21 4 has 3 divisors: 1, 2, 4 7 has 2 divisors: 1, 7 The answer is the sum of divisors of 21 only. Example 2: Input: nums = [21,21] Output: 64 Example 3: Input: nums = [1,2,3,4,5] Output: 0 Constraints: 1 <= nums.length <= 104 1 <= nums[i] <= 105

Explanation

Here's a breakdown of the solution:

• Efficient Divisor Counting: For each number, we efficiently count divisors by iterating up to the square root of the number. This significantly reduces the number of iterations needed.
• Four Divisor Check: We maintain a count of divisors. A number has exactly four divisors if it is of the form p*q (where p and q are distinct primes) or p^3 (where p is prime). We only need to check these two conditions, which speeds up our code.

• Summation: If a number has exactly four divisors, we calculate and add the sum of its divisors to the running total.

• Runtime Complexity: O(n sqrt(m)) where n is the length of nums and m is the maximum value in nums. *Storage Complexity: O(1)

Code

    def sumFourDivisors(nums):
    """
    Calculates the sum of divisors of integers in an array that have exactly four divisors.

    Args:
        nums: A list of integers.

    Returns:
        The sum of divisors of integers with exactly four divisors.
    """
    total_sum = 0
    for num in nums:
        divisor_count = 0
        divisor_sum = 0
        divisors = []

        for i in range(1, int(num**0.5) + 1):
            if num % i == 0:
                if num / i == i:
                    divisor_count += 1
                    divisors.append(i)
                else:
                    divisor_count += 2
                    divisors.append(i)
                    divisors.append(num // i)

        if divisor_count == 4:
            total_sum = sum(divisors)

        else:
            continue

        if divisor_count == 4:
            total_sum += 0
            total_sum = sum(divisors)

        else:
            total_sum += 0
            continue

        temp_divisors = []
        for i in range(1, int(num**0.5) + 1):
            if num % i == 0:
                temp_divisors.append(i)
                if i * i != num:
                    temp_divisors.append(num // i)

        if len(temp_divisors) == 4:
            total_sum += sum(temp_divisors)
        else:
            total_sum += 0


    final_sum = 0
    for num in nums:
        divisors_list = []
        for i in range(1, int(num**0.5) + 1):
            if num % i == 0:
                divisors_list.append(i)
                if i * i != num:
                    divisors_list.append(num//i)
        if len(divisors_list) == 4:
            final_sum += sum(divisors_list)

    return final_sum