<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 tallestBillboard(vector<int>& rods) {
        int n = rods.size();
        int totalSum = accumulate(rods.begin(), rods.end(), 0);
        int limit = 2 * totalSum + 1;

        // Initialize dp arrays for current and previous states
        vector<int> prev(limit, INT_MIN);
        vector<int> curr(limit, INT_MIN);

        // Base case: no rods used, difference is 0
        prev[totalSum] = 0;

        for (int i = 0; i < n; ++i) {
            int rod = rods[i];
            fill(curr.begin(), curr.end(), INT_MIN);  // Reset current state

            for (int j = 0; j < limit; ++j) {
                // Case 1: Add the current rod to one tower
                if (j >= rod && prev[j - rod] != INT_MIN) {
                    curr[j] = max(curr[j], prev[j - rod] + rod);
                }
                
                // Case 2: Add the current rod to the other tower
                if (j + rod < limit && prev[j + rod] != INT_MIN) {
                    curr[j] = max(curr[j], prev[j + rod]);
                }
                
                // Case 3: Ignore the current rod
                if (prev[j] != INT_MIN) {
                    curr[j] = max(curr[j], prev[j]);
                }
            }

            // Move current state to previous state for next iteration
            swap(prev, curr);
        }

        // The result is the maximum height when the difference is 0
        return prev[totalSum];
    }
};

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