# Solving Leetcode Interviews in Seconds with AI: Distinct Prime Factors of Product of Array


	# 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](https://www.chatmagic.app), 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
	```python
	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)
	```