How do you solve Tallest Billboard on LeetCode?

Tallest Billboard (LeetCode #956) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * sum(rods)) time complexity and O(sum(rods)) space complexity. Key insight: The problem can be viewed as a subset-sum problem with constraints.

What is the brute force solution for Tallest Billboard?

The brute force approach for Tallest Billboard is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsets of rods and checking their sums to find pairs that can support the billboard equally. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * sum(rods)).

What algorithm or data structure is used to solve Tallest Billboard?

Tallest Billboard uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Dynamic Programming.

What are the interview tips for Tallest Billboard?

Explain your thought process clearly while coding. Discuss the trade-offs between brute force and optimized solutions. Practice similar DP problems to build confidence.

What are common mistakes when solving Tallest Billboard?

Not considering the absolute difference in heights. Failing to initialize the DP structure correctly.

#956

Tallest Billboard

Hard

ArrayDynamic ProgrammingHash MapDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * sum(rods))
Space	O(1)	O(sum(rods))

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Intuition

Time O(n * sum(rods))Space O(sum(rods))

The optimal solution uses dynamic programming to keep track of the maximum height of the billboard that can be achieved with a given difference in height between the two sides. This reduces the number of combinations we need to check.

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Algorithm

3 steps

1Step 1: Initialize a DP array to store the maximum height for each possible height difference.
2Step 2: Iterate through each rod and update the DP array based on the current rod's height.
3Step 3: At the end, the maximum height for a difference of 0 will give the largest possible height.

solution.py9 lines

1def tallestBillboard(rods):
2    dp = {0: 0}
3    for rod in rods:
4        new_dp = dp.copy()
5        for diff, height in dp.items():
6            new_dp[diff + rod] = max(new_dp.get(diff + rod, 0), height)
7            new_dp[abs(diff - rod)] = max(new_dp.get(abs(diff - rod), 0), height + min(diff, rod))
8        dp = new_dp
9    return dp[0]

ℹ

Complexity note: This complexity comes from iterating through each rod and updating the DP map based on the potential height differences, which can range up to the total sum of the rods.

1The problem can be viewed as a subset-sum problem with constraints.
2Dynamic programming helps in efficiently tracking possible height differences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.