Introduction

In this blog post, we will explore how to solve the LeetCode problem "368" 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 a set of distinct positive integers nums, return the largest subset answer such that every pair (answer[i], answer[j]) of elements in this subset satisfies: answer[i] % answer[j] == 0, or answer[j] % answer[i] == 0 If there are multiple solutions, return any of them. Example 1: Input: nums = [1,2,3] Output: [1,2] Explanation: [1,3] is also accepted. Example 2: Input: nums = [1,2,4,8] Output: [1,2,4,8] Constraints: 1 <= nums.length <= 1000 1 <= nums[i] <= 2 * 109 All the integers in nums are unique.

Explanation

Here's a solution to find the largest divisible subset, along with an explanation:

• Sort the Input: Sorting nums allows us to iterate in ascending order and efficiently check divisibility. This is crucial for building the dynamic programming table.
• Dynamic Programming: We use a DP table dp where dp[i] stores the length of the largest divisible subset ending at nums[i]. We also maintain a predecessor array to reconstruct the actual subset.

• Reconstruction: After filling the DP table, we find the index with the maximum dp value. We then backtrack using the predecessor array to build the largest divisible subset.

• Time & Space Complexity: O(n^2) time, O(n) space

Code

    def largestDivisibleSubset(nums):
    n = len(nums)
    nums.sort()
    dp = [1] * n
    predecessor = [-1] * n
    max_len = 1
    max_index = 0

    for i in range(1, n):
        for j in range(i):
            if nums[i] % nums[j] == 0:
                if dp[i] < dp[j] + 1:
                    dp[i] = dp[j] + 1
                    predecessor[i] = j
        if dp[i] > max_len:
            max_len = dp[i]
            max_index = i

    result = []
    current = max_index
    while current != -1:
        result.append(nums[current])
        current = predecessor[current]

    return result