<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>

cache = {}


# 回傳 nums 列表內，最後一個一定要選的情況下
# 所有元素間都可以彼此整除的最大子集
def lastOneShouldChoose(nums):
    if len(nums) == 1:
        return nums

    key = tuple(nums)

    if key in cache:
        return cache[key]

    maxList = [nums[-1]]  # 至少要有最後一個
    for i in range(len(nums) - 1):
        if nums[-1] % nums[i] == 0:
            l = [*lastOneShouldChoose(nums[0 : i + 1]), nums[-1]]
            if len(l) > len(maxList):
                maxList = l

    cache[key] = maxList
    return maxList


class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        nums.sort()

        maxList = []
        for i in range(len(nums)):  # 每一個都有可能是我的最後一個
            l = lastOneShouldChoose(nums[0 : i + 1])
            if len(l) > len(maxList):
                maxList = l

        return maxList

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