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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2447" 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 greatest common divisor of the subarray's elements is k. A subarray is a contiguous non-empty sequence of elements within an array. The greatest common divisor of an array is the largest integer that evenly divides all the array elements.   Example 1:  Input: nums = [9,3,1,2,6,3], k = 3 Output: 4 Explanation: The subarrays of nums where 3 is the greatest common divisor of all the subarray's elements are: - [9,3,1,2,6,3] - [9,3,1,2,6,3] - [9,3,1,2,6,3] - [9,3,1,2,6,3]  Example 2:  Input: nums = [4], k = 7 Output: 0 Explanation: There are no subarrays of nums where 7 is the greatest common divisor of all the subarray's elements.    Constraints:  1 <= nums.length <= 1000 1 <= nums[i], k <= 109  

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python code:

*   **High-Level Approach:**
    *   Iterate through all possible subarrays of the input array `nums`.
    *   For each subarray, calculate the greatest common divisor (GCD) of its elements.
    *   If the calculated GCD is equal to the target value `k`, increment a counter.

*   **Complexity:**
    *   Runtime Complexity: O(n^2 * log(max(nums[i]))) where n is the length of `nums`.  The n^2 comes from iterating through all subarrays, and the log(max(nums[i])) factor arises from the GCD calculation. In the worst case, `gcd` function may take log(max(a, b)) time.
    *   Storage Complexity: O(1).  The algorithm uses a constant amount of extra space.

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

def subarrayGCD(nums, k):
    """
    Counts the number of subarrays of nums where the greatest common divisor of the subarray's elements is k.
    """
    count = 0
    n = len(nums)

    for i in range(n):
        current_gcd = 0
        for j in range(i, n):
            current_gcd = gcd(current_gcd, nums[j])
            if current_gcd == k:
                count += 1
            elif current_gcd < k:
                break  # Optimization: If GCD becomes less than k, no point continuing

    return count
	```