<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:
    void dfs(int l, int r, int sum, int s, unordered_map<int,int>& mp, vector<int>& rods) {
        if(l == r) {
            mp[sum] = max(mp[sum], s);
            return;
        }
        dfs(l+1, r, sum, s, mp, rods);
        dfs(l+1, r, sum - rods[l], s, mp, rods);
        dfs(l+1, r, sum + rods[l], s + rods[l], mp, rods);
    }
    int tallestBillboard(vector<int>& rods) {
        int n = rods.size();
        unordered_map<int,int> mp1,mp2;
        dfs(0, n/2, 0, 0, mp1, rods);
        dfs(n/2, n, 0, 0, mp2, rods);

        int ans = 0;
        for (auto& [k, v] : mp1) {
            if(mp2.find(-k) != mp2.end()) {
                ans = max(ans, v + mp2[-k]);
            }
        }
        return ans;
    }
};
24253