Introduction

In this blog post, we will explore how to solve the LeetCode problem "3411" 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

You are given an array of positive integers nums. An array arr is called product equivalent if prod(arr) == lcm(arr) * gcd(arr), where: prod(arr) is the product of all elements of arr. gcd(arr) is the GCD of all elements of arr. lcm(arr) is the LCM of all elements of arr. Return the length of the longest product equivalent subarray of nums. Example 1: Input: nums = [1,2,1,2,1,1,1] Output: 5 Explanation: The longest product equivalent subarray is [1, 2, 1, 1, 1], where prod([1, 2, 1, 1, 1]) = 2, gcd([1, 2, 1, 1, 1]) = 1, and lcm([1, 2, 1, 1, 1]) = 2. Example 2: Input: nums = [2,3,4,5,6] Output: 3 Explanation: The longest product equivalent subarray is [3, 4, 5]. Example 3: Input: nums = [1,2,3,1,4,5,1] Output: 5 Constraints: 2 <= nums.length <= 100 1 <= nums[i] <= 10

Explanation

• The core idea is to iterate through all possible subarrays and check if each subarray is product equivalent. The condition prod(arr) == lcm(arr) * gcd(arr) is satisfied if and only if all numbers in the subarray are pairwise relatively prime or if the subarray contains only 1s. We can simplify the check by verifying if the product equals the least common multiple, which holds only under these conditions given positive integers.
  • To efficiently compute the product and least common multiple of a subarray, we can maintain a running product and LCM as we expand the subarray window.
  • Keep track of the longest product equivalent subarray found so far, updating it whenever a longer one is encountered.

• Runtime Complexity: O(n^2), Storage Complexity: O(1)

Code

    def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)

def lcm(a, b):
    return (a * b) // gcd(a, b)

def longest_product_equivalent(nums):
    max_len = 0
    for i in range(len(nums)):
        prod = 1
        lcm_val = 1
        for j in range(i, len(nums)):
            prod *= nums[j]
            lcm_val = lcm(lcm_val, nums[j])

            if prod == lcm_val:
                max_len = max(max_len, j - i + 1)
    return max_len