<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(int idx, int diff, vector<int>& rods) {
        if (idx == rods.size()) {
            if (diff == 0)
                return 0;
            else
                return -1e9;
        }
        int stand1, stand2, reject;
        reject = dfs(idx + 1, diff, rods);
        stand1 = dfs(idx + 1, diff - rods[idx], rods);
        stand2 = rods[idx] + dfs(idx + 1, diff + rods[idx], rods);
        return max({reject, stand1, stand2});
    }
    int memoization(int idx, int diff, vector<int>& rods,
                    vector<vector<int>>& dp) {
        if (idx == rods.size()) {
            if (diff == 0)
                return 0;
            else
                return -1e9;
        }
        if (dp[idx][diff + 5000] != -1)
            return dp[idx][diff + 5000];
        int stand1, stand2, reject;
        reject = memoization(idx + 1, diff, rods, dp);
        stand1 = memoization(idx + 1, diff - rods[idx], rods, dp);
        stand2 = rods[idx] + memoization(idx + 1, diff + rods[idx], rods, dp);
        return dp[idx][diff + 5000] = max({reject, stand1, stand2});
    }
    int tallestBillboard(vector<int>& rods) {
        // return dfs(0, 0, rods);
        vector<vector<int>> dp(25, vector<int>(10000, -1));
        return memoization(0, 0, rods, dp);
    }
};
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