<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>

class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()
        cache = {}
        return self.recursion(0, 1, nums, cache)

    def recursion(self, i, prev, nums, cache):
        if i == len(nums):
            return []

        if (i, prev) in cache:
            return cache[(i, prev)]

        res = self.recursion(i+1, prev, nums, cache)
        if nums[i] % prev == 0:
            temp = [nums[i]] + self.recursion(i+1, nums[i], nums, cache)
            if len(temp) > len(res):
                res = temp
        cache[(i, prev)] = res
        return cache[(i, prev)] 
            

        
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