<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:
    //seperate the stones vector into two subset, and we want this two subsets value as close as possible
    //if the two subset can be equal then we return 0;

    int lastStoneWeightII(vector<int>& stones) {
        if(stones.size()==1){
            return stones[0];
        }
        int sum = accumulate(stones.begin(), stones.end(), 0);
        int min = sum - stones[0];
        int first = 0;
        unordered_map<int,int> m;
        int reserveInt = pow(stones.size()-1, 3);
        m.reserve(reserveInt);
        m.emplace(0,0);
        for(int i = 0; i < stones.size(); i++){
            for(auto j = m.begin(); j != m.end(); j++){
                int first = j->first + stones[i];
                int mo =  abs(first - (sum - first));
                if(min > mo){
                    min = mo;
                }
                if(m.count(first) <= 0 && first > 0){
                    m.emplace(first, min);
                }
            }
        }
        return min;
    }
};
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