Tallest Billboard
<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>
Array; Dynamic Programming
class Solution {
public:
    int tallestBillboard(vector<int>& rods) {
        int s = accumulate(rods.begin(), rods.end(), 0);
        int n = rods.size();
        int f[n][s + 1];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (i >= n) {
                return j == 0 ? 0 : -(1 << 30);
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            int ans = max(dfs(i + 1, j), dfs(i + 1, j + rods[i]));
            ans = max(ans, dfs(i + 1, abs(j - rods[i])) + min(j, rods[i]));
            return f[i][j] = ans;
        };
        return dfs(0, 0);
    }
};