<p>Given a set of <strong>distinct</strong> positive integers <code>nums</code>, return the largest subset <code>answer</code> such that every pair <code>(answer[i], answer[j])</code> of elements in this subset satisfies:</p>

<ul>
	<li><code>answer[i] % answer[j] == 0</code>, or</li>
	<li><code>answer[j] % answer[i] == 0</code></li>
</ul>

<p>If there are multiple solutions, return any of them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [1,2]
<strong>Explanation:</strong> [1,3] is also accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,4,8]
<strong>Output:</strong> [1,2,4,8]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2 * 10<sup>9</sup></code></li>
	<li>All the integers in <code>nums</code> are <strong>unique</strong>.</li>
</ul>

from typing import List

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        @cache
        def helper(prev, curr):
            # Base case: when we've considered all elements
            if curr == len(nums):
                return []

            # Option 1: Include nums[curr] if it's divisible by nums[prev]
            take = []
            if prev == -1 or nums[curr] % nums[prev] == 0:
                take = helper(curr, curr + 1)  # Continue from next index
                take = [nums[curr]] + take  # Include nums[curr] in the subset

            # Option 2: Skip nums[curr]
            skip = helper(prev, curr + 1)

            # Return the larger subset
            if len(take) > len(skip):
                return take
            else:
                return skip

        # Sort to ensure divisibility condition
        nums.sort()
        return helper(-1, 0)

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