<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) {
        int n = rods.size();
        int sum = accumulate(rods.begin(), rods.end(), 0);
        vector<int> dp(2 * sum + 1, -1); // dp[i] is maximum left rod when left - right = i, [-sum:sum] -> [0:2sum]
        dp[sum] = 0;
        
        for (auto& rod : rods) {
            auto dp_old = dp;
            for (int d = -sum; d <= sum; d++) {
                if (dp_old[d + sum] == -1)
                    continue;
                
                if (d + rod <= sum) // add to left
                    dp[d + rod + sum] = max(dp[d + rod + sum], dp_old[d + sum] + rod);
                if (d - rod >= -sum) // add to right
                    dp[d - rod + sum] = max(dp[d - rod + sum], dp_old[d + sum]);
            }
        }
        
        return dp[sum];
    }
};
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