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, 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

    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