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, 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

    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