Introduction

In this blog post, we will explore how to solve the LeetCode problem "2521" 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 array of positive integers nums, return the number of distinct prime factors in the product of the elements of nums. Note that: A number greater than 1 is called prime if it is divisible by only 1 and itself. An integer val1 is a factor of another integer val2 if val2 / val1 is an integer. Example 1: Input: nums = [2,4,3,7,10,6] Output: 4 Explanation: The product of all the elements in nums is: 2 4 3 7 10 6 = 10080 = 25 32 5 7. There are 4 distinct prime factors so we return 4. Example 2: Input: nums = [2,4,8,16] Output: 1 Explanation: The product of all the elements in nums is: 2 4 8 * 16 = 1024 = 210. There is 1 distinct prime factor so we return 1. Constraints: 1 <= nums.length <= 104 2 <= nums[i] <= 1000

Explanation

Here's the approach to solve this problem:

• Prime Factorization: Iterate through the input array nums. For each number, find its prime factors.
• Distinct Set: Store the distinct prime factors in a set to avoid duplicates.

• Return Count: Return the size of the set, which represents the number of distinct prime factors.

• Runtime Complexity: O(n sqrt(m)), where n is the length of nums and m is the maximum value in nums. *Storage Complexity: O(k), where k is the number of distinct prime factors (at most log m).

Code

    def distinctPrimeFactors(nums):
    """
    Calculates the number of distinct prime factors in the product of the elements of nums.

    Args:
        nums: A list of positive integers.

    Returns:
        The number of distinct prime factors in the product of nums.
    """

    distinct_primes = set()

    for num in nums:
        # Find prime factors of num
        i = 2
        temp = num
        while i * i <= temp:
            if temp % i == 0:
                distinct_primes.add(i)
                while temp % i == 0:
                    temp //= i
            i += 1
        if temp > 1:
            distinct_primes.add(temp)

    return len(distinct_primes)