<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 {

public:

    int lastStoneWeightII(vector<int>& stones) {
        vector<unordered_map<int, int>> cache (stones.size());
        return solve(stones, 0, 0, cache);
    }

    int solve(vector<int> & stones, int running_sum, int index, vector<unordered_map<int, int>> & cache)
    {
        if (cache[index].contains(running_sum))
        {
            return cache[index][running_sum];
        }

        if (index == stones.size() - 1)
        {
            int s1 = running_sum - stones[index];
            cache[index][running_sum] = max(s1, -s1);

            return cache[index][running_sum];
        }

        int v1 = solve(stones, running_sum - stones[index], index + 1, cache);
        int v2 = solve(stones, stones[index] - running_sum, index + 1, cache);

        cache[index][running_sum] = min(v1, v2);
        return min(v1, v2);
    }
};
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