Introduction

In this blog post, we will explore how to solve the LeetCode problem "1819" 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 nums that consists of positive integers. The GCD of a sequence of numbers is defined as the greatest integer that divides all the numbers in the sequence evenly. For example, the GCD of the sequence [4,6,16] is 2. A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array. For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10]. Return the number of different GCDs among all non-empty subsequences of nums. Example 1: Input: nums = [6,10,3] Output: 5 Explanation: The figure shows all the non-empty subsequences and their GCDs. The different GCDs are 6, 10, 3, 2, and 1. Example 2: Input: nums = [5,15,40,5,6] Output: 7 Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 2 * 105

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: Iterate through all possible GCDs from 1 up to the maximum value in nums. For each potential GCD g, check if it can be formed by a subsequence of nums. A GCD g can be formed if all numbers in the subsequence are divisible by g, and there exists at least one number in the subsequence such that g is its only factor from nums. This ensures we consider only actual GCDs that can be formed.

• Optimization: We efficiently check if a GCD is possible by iterating through multiples of the potential GCD in the input array. If we find at least one multiple, we compute the GCD of all multiples found in the array. If the result equals the potential GCD, we know that it's a valid GCD for a subsequence.

• Complexity:

  • Runtime: O(M log M) where M is the maximum element in nums.
  • Space: O(1)

Code

    from math import gcd

def countDifferentGCDs(nums):
    max_num = max(nums)
    present = [False] * (max_num + 1)
    for num in nums:
        present[num] = True

    count = 0
    for i in range(1, max_num + 1):
        curr_gcd = 0
        found = False
        for j in range(i, max_num + 1, i):
            if present[j]:
                curr_gcd = gcd(curr_gcd, j)
                found = True

        if found and curr_gcd == i:
            count += 1

    return count