# Solving Leetcode Interviews in Seconds with AI: Number of Subarrays With LCM Equal to K


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2470" 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 integer array nums and an integer k, return the number of subarrays of nums where the least common multiple of the subarray's elements is k. A subarray is a contiguous non-empty sequence of elements within an array. The least common multiple of an array is the smallest positive integer that is divisible by all the array elements.   Example 1:  Input: nums = [3,6,2,7,1], k = 6 Output: 4 Explanation: The subarrays of nums where 6 is the least common multiple of all the subarray's elements are: - [3,6,2,7,1] - [3,6,2,7,1] - [3,6,2,7,1] - [3,6,2,7,1]  Example 2:  Input: nums = [3], k = 2 Output: 0 Explanation: There are no subarrays of nums where 2 is the least common multiple of all the subarray's elements.    Constraints:  1 <= nums.length <= 1000 1 <= nums[i], k <= 1000  

	# Explanation
	*   **Iterate through all possible subarrays:** The solution iterates through all possible subarrays of the input array `nums` using nested loops.
*   **Calculate the LCM of each subarray:** For each subarray, it calculates the least common multiple (LCM) of its elements.
*   **Count subarrays with LCM equal to k:** If the calculated LCM is equal to the given integer `k`, the count is incremented.

*   **Runtime Complexity:** O(n^2 * log(m)), where n is the length of nums and m is the maximum element in nums.
    **Storage Complexity:** O(1)

	
	# Code
	```python
	import math

def gcd(a, b):
    """
    Calculates the greatest common divisor (GCD) of two integers using the Euclidean algorithm.
    """
    while b:
        a, b = b, a % b
    return a

def lcm(a, b):
    """
    Calculates the least common multiple (LCM) of two integers.
    """
    if a == 0 or b == 0:
      return 0
    return (a * b) // gcd(a, b)

def subarrayLCM(nums, k):
    """
    Counts the number of subarrays in nums where the LCM of the subarray's elements is k.
    """
    count = 0
    n = len(nums)

    for i in range(n):
        current_lcm = 1
        for j in range(i, n):
            current_lcm = lcm(current_lcm, nums[j])
            if current_lcm == k:
                count += 1
            elif current_lcm > k:
                break

    return count
	```