<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 solve(int ind, int val, int size, vector<int>& stones,
              vector<vector<int>>& dp) {
        if (ind == size) {
            if (val < 0)
                return 1e9;
            return dp[ind][val] = 0;
        }
        if (dp[ind][val + 3000] != -1)
            return dp[ind][val + 3000];
        int pos =
            stones[ind] + solve(ind + 1, val + stones[ind], size, stones, dp);
        int neg =
            -stones[ind] + solve(ind + 1, val - stones[ind], size, stones, dp);
        return dp[ind][val + 3000] = min(pos, neg);
    }
    int lastStoneWeightII(vector<int>& stones) {
        int n = stones.size();
        int sm = accumulate(stones.begin(), stones.end(), 0);
        vector<vector<int>> dp(n + 1, vector<int>(6000, -1));
        return solve(0, 0, n, stones, dp);
    }
};
33687