<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:

    // void solve(vector<int> stones, int i, int& ans) {
    //     //if(i == stones.size() - 1) ans = min(ans, stones[i]);
        
    //     bool flag = false;
    //     for(int j = i + 1; j < stones.size(); j++) {
    //         if(stones[j] == 0) continue;
    //         vector<int> temp = stones;
    //         temp[j] = abs(stones[j] - stones[i]);
    //         flag = true;
    //         solve(temp, i + 1, ans);
    //     }
    //     if(flag == false) ans = min(ans, stones[i]);
    // }

    int solve(vector<int>& stones, int i, int j, vector<vector<int>>& dp, int& sum) {
        if(j >= ceil(sum/2) || i == stones.size()) {
            return abs(j - (sum - j));
        }
        if(dp[i][j] != -1) return dp[i][j];
        int a = solve(stones, i + 1, j, dp, sum);
        int b = solve(stones, i + 1, j + stones[i], dp, sum);
        dp[i][j] = min(a, b);
        return dp[i][j];
    }

    int lastStoneWeightII(vector<int>& stones) {
        // int ans = INT_MAX;
        
        // solve(stones, 0, ans);
        // return ans;
        int sum = accumulate(stones.begin(), stones.end(), 0);
        vector<vector<int>> dp(stones.size(), vector<int> (3000, -1));
        return solve(stones, 0, 0, dp, sum);
    }
};
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