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 {
    private Integer[][] f;
    private int[] rods;
    private int n;

    public int tallestBillboard(int[] rods) {
        int s = 0;
        for (int x : rods) {
            s += x;
        }
        n = rods.length;
        this.rods = rods;
        f = new Integer[n][s + 1];
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= n) {
            return j == 0 ? 0 : -(1 << 30);
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = Math.max(dfs(i + 1, j), dfs(i + 1, j + rods[i]));
        ans = Math.max(ans, dfs(i + 1, Math.abs(rods[i] - j)) + Math.min(j, rods[i]));
        return f[i][j] = ans;
    }
}