<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 n;
    long long total;
    vector<vector<long long>>dp;
   
   // 3d to 2d dp conversion is necessary


    long long f(int i,int diff,vector<int>&rods)
    {
        if(i>=n)
        {
            if(diff==0){
                cout<<"yy"<<endl;
            return 0;

            }

            return LLONG_MIN;
        }

     

        if(dp[i][diff+5000]!=-1)
        {
            return dp[i][diff+5000];
        }


       long long result=LLONG_MIN;
       long long case1=f(i+1,diff,rods);
       long long case2=rods[i]+f(i+1,diff+rods[i],rods);
       long long case3=rods[i]+f(i+1,diff-rods[i],rods);
    //    cout<<i<<" "<<case1<<" "<<case2<<" "<<case3<<endl;
       result=max(result,case1);
       result=max(result,case2);
       result=max(result,case3);

       return dp[i][diff+5000]=result;
    }

    int tallestBillboard(vector<int>& rods) 
    {
        n=rods.size();
        long long sum=0;
        dp=vector<vector<long long>>(n+1,vector<long long>(10003,-1));
        long long result=f(0,0,rods)/2;
        return result;
    }
};
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