<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]:
        #1 2 3
        # mod is 0 --> exactly divisible
        # or we can say that nums[i] = k * nums[j] 
            # k is an integer
        
        # 1,2
            # 1,3
                # 2,3  
        # 3,1
            # 2
        # 3,2
            # 1
        if len(nums) < 2: return nums
        nums.sort() # next comparison
        cache = {}
        def rec(i=0,prev=1):
            if i == len(nums):
                return []
            if (i,prev) in cache: return cache[(i,prev)]
            
            # exclude this pair
            res = rec(i+1,prev)
            # include
            if nums[i] % prev == 0:
                res = max( [nums[i]]+rec(i+1,nums[i]), res ,key=len)
            cache[(i,prev)] = res
            return res
        return rec()
             

        
        
        
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