<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:
        """
        Imagine you have 5 stones: a, b, c, d, e
        You want to minimize ((b-a) - (c - d)) - e = x, where the rocks a through e can come in any order to make this possible
        ((b-a) - (c - d)) - e = 0, since 0 is the smallest value you can possibly have that's the minimum value we are going to set for x for now
        b - a - c + d - e = 0
        b + d = a + c + e
        So we basically want to find a way to partition the numbers in a way so that they are as equal as possible. The difference of these partitions is the answer
        """
        # BRUTE FORCE: O(2^N) time O(N) space. We can choose to include or not include every value in stones to try to get a partition that is as close to
        # the sum of sum(stones) / 2 as possible. Then we can find the bigger half's value by doing sum(stones) - smallerPartition (which was obtained from recursion)
        # Finally, return biggerPartition - smallerPartition to get the answer
        # MEMOIZATION: O(N * sum(stones) / 2) time and space. Use a cache to memoize every possible combination of the iterator i and cumulative sum curr
        total = sum(stones)
        target = total / 2
        dp = dict()
        def solve(stones, i, curr, goal):
            if curr > goal:
                return 0 # We passed the goal so return 0
            if i >= len(stones):
                return curr
            if i in dp and curr in dp[i]:
                return dp[i][curr]
            skip = solve(stones, i + 1, curr, goal)
            include = solve(stones, i + 1, curr + stones[i], goal)
            if i not in dp:
                dp[i] = dict()
            dp[i][curr] = max(skip, include)
            return dp[i][curr]

        smaller_partition = solve(stones, 0, 0, target)
        bigger_partition = total - smaller_partition
        return bigger_partition - smaller_partition
19900