Formatted question description: https://leetcode.ca/all/956.html

956. Tallest Billboard (Hard)

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 have a collection of rods which 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:

Input: [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:

Input: [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:

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

 

Note:

  1. 0 <= rods.length <= 20
  2. 1 <= rods[i] <= 1000
  3. The sum of rods is at most 5000.

Related Topics:
Dynamic Programming

Solution 1. DP

Thought

For each rod x, we have 3 options:

  1. use it in one post
  2. use it in another post
  3. don’t use it.

If we turn all the numbers used in the first post to negative (turn x to -x), leave the numbers used in the second post as-is (x to +x), and turn all the numbers that are not used to 0, this question turns into:

Find the max score we can get after doing the above operations. The “score” is the sum of all the positive numbers. For example, +1 +2 +3 -6 has a score of 6.

Since sum(rods) is bounded, it suggests us to use this fact in some way.

A fact we should consider is that for a given sum, it doesn’t matter how we get the sum.

For example, with rods = [1,2,2,3], we could get sum 3 in 3 different ways. If we just consider sum = 3, we actually covered all those three cases.

Since sum is in range [-5000, 5000], we just have 10001 numbers to consider.

Algorithm

Let dp[i][s] be the largest score we can get using rods[(i+1)..(N-1)] to get sum s.

For example, for rods = [1,2,3,6], we have dp[1][1] = 5, because after writing 1, we could write +2 +3 -6 to get sum 1, and the corresponding score is 5.

For the base case, dp[rods.length][s] is 0 when s == 0, and -infinity everywhere else.

The recursion is dp[i][s] = max(dp[i+1][s], dp[i+1][s-rods[i]], rods[i] + dp[i+1][s+rods[i]]).

NOTE: in the following implementation we use sum = 5000 as sum = 0 to simply code.

// OJ: https://leetcode.com/problems/tallest-billboard

// Time: O(NS) where N is the length of `rods`,
//             and S is the maximum of `sum(rods[i..j])`
// Space: O(NS)
// Ref: https://leetcode.com/articles/tallest-billboard/
class Solution {
private:
    vector<vector<int>> dp;
    int dfs(vector<int>& rods, int i, int s) {
        if (i == rods.size()) return s == 5000 ? 0 : INT_MIN;
        if (dp[i][s] != INT_MIN) return dp[i][s];
        int ans = dfs(rods, i + 1, s);
        ans = max(ans, dfs(rods, i + 1, s - rods[i]));
        ans = max(ans, rods[i] + dfs(rods, i + 1, s + rods[i]));
        return dp[i][s] = ans;
    }
public:
    int tallestBillboard(vector<int>& rods) {
        int N = rods.size();
        dp = vector<vector<int>>(N, vector<int>(10001, INT_MIN));
        return dfs(rods, 0, 5000);
    }
};

Java

class Solution {
    public int tallestBillboard(int[] rods) {
        int rows = rods.length;
        if (rows < 2)
            return 0;
        int sum = 0;
        for (int rod : rods)
            sum += rod;
        int columns = sum * 2 + 1;
        int[][] dp = new int[rows][columns];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++)
                dp[i][j] = -1;
        }
        dp[0][sum + rods[0]] = rods[0];
        dp[0][sum] = 0;
        dp[0][sum - rods[0]] = 0;
        for (int i = 1; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (dp[i - 1][j] >= 0) {
                    int rod = rods[i];
                    dp[i][j + rod] = Math.max(dp[i][j + rod], dp[i - 1][j] + rod);
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]);
                    dp[i][j - rod] = Math.max(dp[i][j - rod], dp[i - 1][j]);
                }
            }
        }
        int maxHeight = 0;
        for (int i = 0; i < rows; i++)
            maxHeight = Math.max(maxHeight, dp[i][sum]);
        return maxHeight;
    }
}

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