<p>You are given an array of integers <code>stones</code> where <code>stones[i]</code> is the weight of the <code>i<sup>th</sup></code> stone.</p>

<p>We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights <code>x</code> and <code>y</code> with <code>x &lt;= y</code>. The result of this smash is:</p>

<ul>
	<li>If <code>x == y</code>, both stones are destroyed, and</li>
	<li>If <code>x != y</code>, the stone of weight <code>x</code> is destroyed, and the stone of weight <code>y</code> has new weight <code>y - x</code>.</li>
</ul>

<p>At the end of the game, there is <strong>at most one</strong> stone left.</p>

<p>Return <em>the smallest possible weight of the left stone</em>. If there are no stones left, return <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> stones = [2,7,4,1,8,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that&#39;s the optimal value.
</pre>

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

<pre>
<strong>Input:</strong> stones = [31,26,33,21,40]
<strong>Output:</strong> 5
</pre>

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

<ul>
	<li><code>1 &lt;= stones.length &lt;= 30</code></li>
	<li><code>1 &lt;= stones[i] &lt;= 100</code></li>
</ul>

class Solution:
    
    def lastStoneWeightII(self, stones: List[int]) -> int:
        self.total = sum(stones)
        self.target = self.total // 2
        self.smallest_diff = math.inf
        self.stones = stones
        self.cache = defaultdict(bool)

        self.find_smallest_diff(0, 0)


        a_pile = self.total - (self.target - self.smallest_diff)
        b_pile = self.total - a_pile 

        return abs(a_pile - b_pile)

    def find_smallest_diff(self, stone: int, sum_so_far: int): 
        if (stone, sum_so_far) in self.cache: 
            return 
            
        self.smallest_diff = min(self.smallest_diff, abs(sum_so_far - self.target))
        if sum_so_far > self.target: 
            return 
        
        self.cache[(stone, sum_so_far)] = True

        if stone == len(self.stones): 
            return 

        self.find_smallest_diff(stone + 1, sum_so_far)
        self.find_smallest_diff(stone + 1, sum_so_far + self.stones[stone])


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