<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 dfs(vector<int>& rs, int i, int s1, int s2, int m_sum, vector<vector<int>> &m) {
        if (s1 > m_sum || s2 > m_sum) return -1;
        if (i >= rs.size()) return s1 == s2 ? s1 : -1;
        if (m[i][abs(s1 - s2)] == -2) {
            m[i][abs(s1 - s2)] = max(-1, max({dfs(rs, i + 1, s1, s2, m_sum, m), dfs(rs, i + 1, s1 + rs[i], s2, m_sum, m), dfs(rs, i + 1, s1, s2 + rs[i], m_sum, m)}) - max(s1, s2));
        }
        return m[i][abs(s1 - s2)] + (m[i][abs(s1 - s2)] == -1 ? 0 : max(s1, s2));
    }

    int tallestBillboard(vector<int>& rods) {
        int m_sum = accumulate(begin(rods), end(rods), 0) / 2;
        vector<vector<int>> m(rods.size(), vector<int>(m_sum + 1, -2));
        return max(0, dfs(rods, 0, 0, 0, m_sum, m));
    }
};

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