<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>

/*You are given an array of integers stones where stones[i] is the weight of the ith stone.
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 x and y with x <= y. The result of this smash is:
If x == y, both stones are destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.
At the end of the game, there is at most one stone left.
Return the smallest possible weight of the left stone. If there are no stones left, return 0.
*/
class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        /*  1 <= stones.length <= 30
            1 <= stones[i] <= 100
        */
        unordered_set<int> set({0});
        for(auto x:stones) {
            auto set_temp = set;
            set.clear();
            for(auto s:set_temp) {
                set.insert(s+x);
                set.insert(s-x);
            };
        }

        int res = INT_MAX;
        for(auto x:set) {
            if(x>=0 && x<res)
                res = x;
        }
        return res;
    }
};
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