# Tallest Billboard Problem

## Description

LeetCode Problem 956.

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.

You are given a collection of rods that can be welded together. For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6.

Return the largest possible height of your billboard installation. If you cannot support the billboard, return 0.

Example 1:

``````1
2
3
Input: rods = [1,2,3,6]
Output: 6
Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.
``````

Example 2:

``````1
2
3
Input: rods = [1,2,3,4,5,6]
Output: 10
Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.
``````

Example 3:

``````1
2
3
Input: rods = [1,2]
Output: 0
Explanation: The billboard cannot be supported, so we return 0.
``````

Constraints:

• 1 <= rods.length <= 20
• 1 <= rods[i] <= 1000
• sum(rods[i]) <= 5000

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
public:
int tallestBillboard(vector<int>& nums) {
int sum = accumulate(nums.begin(), nums.end(), 0);
vector<int> dp(sum + 1, -1), curr;
dp[0] = 0;
for (int &el : nums) {
// storing the immediate calculated values
curr = dp;
for (int i = 0; i <= sum - el; ++i) {
// not computed still now, so skip it
if (curr[i] < 0)
continue;
// putting rod into heighest side
dp[i + el] = max(dp[i + el], curr[i]);
// putting rod into smallest side
dp[abs(i - el)] = max(dp[abs(i - el)], curr[i] + min(i, el));
}
}
// we need equal sizes so diff =0 will be my ans
return dp[0];
}
};
``````