<p>You are installing a billboard and want it to have the largest height. The billboard will have two steel supports, one on each side. Each steel support must be an equal height.</p>

<p>You are given a collection of <code>rods</code> that can be welded together. For example, if you have rods of lengths <code>1</code>, <code>2</code>, and <code>3</code>, you can weld them together to make a support of length <code>6</code>.</p>

<p>Return <em>the largest possible height of your billboard installation</em>. If you cannot support the billboard, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> rods = [1,2,3,6]
<strong>Output:</strong> 6
<strong>Explanation:</strong> We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> rods = [1,2,3,4,5,6]
<strong>Output:</strong> 10
<strong>Explanation:</strong> We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> rods = [1,2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The billboard cannot be supported, so we return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= rods.length &lt;= 20</code></li>
	<li><code>1 &lt;= rods[i] &lt;= 1000</code></li>
	<li><code>sum(rods[i]) &lt;= 5000</code></li>
</ul>

class Solution {
public:
    int tallestBillboard(vector<int>& rods) {
        /*
            split to two group
            
            x + y <= sum
            x == y

            dp[diff] = max left subset sum with left subset sum - right subset sum = diff
            
            left_sum - right_sum = diff
            1. (left_sum + x) - right_sum = diff+x
            2. left_sum - (right_sum + x) = diff-x
            dp[diff+x] = dp[diff] + x
            dp[diff-x] = dp[diff]
        */
        int n = rods.size();
        int offset = 5000;
        vector<int> dp(offset*2+1, -1);
        dp[offset] = 0;

        for(auto x: rods){
            auto prev_dp = dp;
            for(auto diff=-offset; diff<=offset; diff++){
                int j = diff + offset;
                if(prev_dp[j]==-1) continue;

                if(diff+x<=offset) dp[j+x] = max(dp[j+x], prev_dp[j] + x);
                if(diff-x>=-offset) dp[j-x] = max(dp[j-x], prev_dp[j]);
            }
        }
        return dp[offset];
    }
};
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