<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 dp(vector<int>& rods, int difference, int i, vector<vector<int>>& mem){
        if(i == rods.size()){
            if(difference == 0){
                return 0;
            }
            return INT_MIN;
        }
        if(mem[difference + 5000][i] == -1){
            long l = dp(rods, difference + rods[i], i + 1, mem);
            long r = dp(rods, difference - rods[i], i + 1, mem);
            long n = dp(rods, difference, i + 1, mem);
            int m = max(l + rods[i], max(r, n));
            mem[difference + 5000][i] = m;
        }
        return mem[difference + 5000][i];

    }
    int tallestBillboard(vector<int>& rods) {
        int sum = 0;
        for(int i = 0; i < rods.size(); i++){
            sum += rods[i];
        }
        vector<vector<int>> mem(10000, vector<int>(rods.size(), -1));
        return dp(rods, 0, 0, mem);
    }
};
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