<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:
    def tallestBillboard(self, rods: List[int]) -> int:
        n = len(rods)
        # dp[i][d] = dp[i - 1][d], dp[i - 1][d + rods[i]] + rods[i], dp[i - 1][d - rods[i]] + rods[i]

        sm = sum(rods)

        # @cache
        # def dfs(i, d):
        #     if i == 0:
        #         if rods[i] == d:
        #             return d
        #         elif d == 0:
        #             return 0
        #         else:
        #             return -inf
        #     ans = dfs(i - 1, d)
        #     ans = max(ans, dfs(i - 1, d + rods[i]) + rods[i])
        #     ans = max(ans, dfs(i - 1, abs(d - rods[i])) + rods[i])
        #     return ans
        # t = dfs(n - 1,0)
        # if t <= 0:
        #     return 0
        # else:
        #     return t // 2
        dp = [[-inf] * (sm + 1) for _ in range(n + 1)]
        # dp = [-inf] * (sm + 1)
        dp[0][0]= 0
        
        for i in range(n):
            for j in range(sm - rods[i] + 1):
                dp[i + 1][j] = max(dp[i][j], dp[i][abs(j - rods[i])] + rods[i], dp[i][j + rods[i]] + rods[i])
        
        t = dp[n][0]
        if t == -inf:
            return 0
        else:
            return t // 2
        
        
        
        
18000