<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]:
        self.dp = {}
        nums.sort()
        return self.dfs(nums, 0, None)

    def dfs(self, nums, i, prev):

        if(i == len(nums)):
            return []
        if((i, prev) in self.dp):
            return self.dp[(i, prev)]
        
        r1 = []
        if(prev is None or (prev!=None and nums[i]%nums[prev]==0)):
            # res.append(nums[i])
            r1 = [nums[i]] + self.dfs(nums, i+1, i)
            # res.pop()

        r2 = self.dfs(nums, i+1, prev)

        if(len(r1) > len(r2)):
            self.dp[(i, prev)] = r1
            return r1
        
        self.dp[(i, prev)] = r2
        return r2
        
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