#956

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Tallest Billboard

Solve the Tallest Billboard problem by using dynamic programming to find the maximum equal height for two disjoint rod subsets.

array dynamic programming

Problem Statement

You are tasked with building a billboard that requires two steel supports. These supports must have equal heights, and you have a collection of rods that can be welded together to form the supports. The goal is to determine the largest possible height for these supports, such that the total height of both supports is the same.

Given an array of rod lengths, the task is to find two disjoint subsets of rods that have equal total lengths. If this is impossible, return 0. The solution needs to efficiently find this balance using dynamic programming to ensure that the two supports are of equal height while maximizing the height of the billboard.

Examples

Example 1

Input: rods = [1,2,3,6]

Output: 6

We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.

Example 2

Input: rods = [1,2,3,4,5,6]

Output: 10

We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.

Example 3

Input: rods = [1,2]

Output: 0

The billboard cannot be supported, so we return 0.

Constraints

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

Solution Approach

State Transition Dynamic Programming

Use dynamic programming (DP) to track possible sums that can be achieved by selecting subsets of rods. Maintain a DP table to represent different possible balances between the two supports. For each rod, update the table to reflect the new possible sums that can be achieved, considering both inclusion and exclusion of the rod.

Subset Sum Approach

The problem can be viewed as a variant of the subset sum problem, where we attempt to divide the rods into two equal subsets. By exploring all possible combinations of rods, we check if there’s a way to partition the rods such that both supports are of equal height, maximizing this height.

Optimize Space Complexity

Since the problem's space complexity can be reduced, maintain a compact DP array that only tracks necessary states rather than keeping track of all potential states. This will optimize the memory usage and make the solution more efficient.

Complexity Analysis

Metric	Value
Time	O(n\cdot m)
Space	O(m)

The time complexity of this problem is O(n⋅m), where n is the number of rods and m is the target sum (half of the total rod length). The space complexity is O(m) as the solution uses a dynamic programming table to store possible sums. Both complexities ensure the problem can be solved efficiently given the constraints.

What Interviewers Usually Probe

Candidate should demonstrate an understanding of dynamic programming and subset sum problems.
Look for clarity in explaining the state transition logic and how the DP table is updated.
The candidate should optimize the solution in terms of both time and space complexity.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the requirement of two disjoint subsets; candidate might focus only on a single subset sum.
Failing to properly optimize space complexity, resulting in an inefficient solution.
Not correctly updating the DP table or missing necessary states, which could lead to an incorrect answer.

Follow-up variants

Consider a variation where rod lengths are constrained to a specific range.
Extend the problem to handle more complex rod configurations or additional constraints.
Modify the problem to work in a scenario where rod lengths must follow a specific pattern or sequence.

FAQ

How do I solve the Tallest Billboard problem using dynamic programming?

The Tallest Billboard problem can be solved by using dynamic programming to keep track of possible sums of rods for the two supports, updating a DP table with each rod.

What is the main challenge in solving Tallest Billboard?

The key challenge is ensuring the correct partitioning of rods into two subsets with equal sum while using dynamic programming efficiently to handle the space and time constraints.

What are the space and time complexities of the Tallest Billboard problem?

The time complexity is O(n⋅m) and the space complexity is O(m), where n is the number of rods and m is the total sum of rods divided by two.

What are the common pitfalls in solving Tallest Billboard?

Common pitfalls include misunderstanding the requirement for two disjoint subsets, failing to optimize the DP table, or incorrectly managing the states.

How can GhostInterview assist in solving the Tallest Billboard problem?

GhostInterview helps by providing guidance on dynamic programming techniques, offering practice on the specific challenges of the problem, and ensuring optimization of time and space complexity.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def tallestBillboard(self, rods: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= len(rods):
                return 0 if j == 0 else -inf
            ans = max(dfs(i + 1, j), dfs(i + 1, j + rods[i]))
            ans = max(ans, dfs(i + 1, abs(rods[i] - j)) + min(j, rods[i]))
            return ans

        return dfs(0, 0)

Solution 2

#### Python3

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class Solution:
    def tallestBillboard(self, rods: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= len(rods):
                return 0 if j == 0 else -inf
            ans = max(dfs(i + 1, j), dfs(i + 1, j + rods[i]))
            ans = max(ans, dfs(i + 1, abs(rods[i] - j)) + min(j, rods[i]))
            return ans

        return dfs(0, 0)

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