What is the time complexity of Tallest Billboard?

The optimal solution has a time complexity of O(n*sum(rods)).

What is the space complexity of Tallest Billboard?

The space complexity is O(sum(rods)).

What data structures are used to solve Tallest Billboard?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Tallest Billboard?

Insight 1: The problem can be modeled as finding two subsets of rods with equal sums. This is a variation of the subset sum problem. Insight 2: Dynamic programming is an effective approach for solving this problem. We can use a map to store the difference between the sums of the two subsets and the corresponding maximum sum of one of the subsets.

Which companies ask LeetCode 956. Tallest Billboard in interviews?

This problem is frequently asked at Atlassian.

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Tallest Billboard - LeetCode 956 Solution

Tallest Billboard - Complete Solution Guide

Tallest Billboard is LeetCode problem 956, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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: Input: rods = [1

Detailed Explanation

The "Tallest Billboard" problem asks us to find the maximum possible height of a billboard that can be supported by two steel supports of equal height. We are given a collection of rods with varying lengths. We need to divide these rods into two disjoint subsets such that the sum of the rods in each subset is the same. This common sum represents the height of each steel support, and therefore the height of the billboard. If it is not possible to create two such subsets, the answer is 0.

Solution Approach

The solution utilizes dynamic programming to efficiently find the maximum possible height. We use a map (dictionary in Python) `dp` where the key is the difference between the sums of the two partitions (left and right supports), and the value is the maximum sum achieved for the left partition (sum1) with that difference. We iterate through the `rods` array, updating the `dp` map in each step. For each rod, we consider adding it to the left partition (increasing the difference), or to the right partition (decreasing the difference). We update the `dp` map with the new differences and corresponding sums. The algorithm efficiently explores all possible combinations of rods to find the maximum possible height.

Step-by-Step Algorithm

Step 1: Initialize a map `dp` to store the differences between the sums of the two partitions and the corresponding maximum sum of one of the subsets. Initialize `dp[0] = 0` because initially, the difference is zero and the sum is also zero.
Step 2: Iterate through each `rod` in the `rods` array.
Step 3: Create a copy `cur` of the `dp` map to avoid modifying the map while iterating over it.
Step 4: Iterate through the `cur` map. For each entry (difference, sum1) in `cur`:
Step 5: Add the current `rod` to the first partition. Update `dp[difference + rod]` with the maximum of the current value and `sum1 + rod`.
Step 6: Add the current `rod` to the second partition. Update `dp[difference - rod]` with the maximum of the current value and `sum1`.
Step 7: After processing all the rods, return `dp[0]` which contains the maximum possible height when the difference between the two partitions is 0. If `dp[0]` does not exist return 0.
Step 8: In the C implementation, a custom hash table is utilized to perform dynamic programming due to the lack of built in hash map functionality. It also incorporates memory management, and ensures the proper initialization of entries to differentiate them from present values.

Key Insights

Insight 1: The problem can be modeled as finding two subsets of rods with equal sums. This is a variation of the subset sum problem.
Insight 2: Dynamic programming is an effective approach for solving this problem. We can use a map to store the difference between the sums of the two subsets and the corresponding maximum sum of one of the subsets.
Insight 3: The difference between the sums of the two subsets can range from -sum(rods) to +sum(rods). Thus, we need to keep track of all possible differences. The key in the dp map represents the difference.

Complexity Analysis

Time Complexity: O(n*sum(rods))

Space Complexity: O(sum(rods))

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Atlassian.

Tallest Billboard | LeetCode 956

Hard

Array Dynamic Programming

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int tallestBillboard(int[] rods) {
        Map<Integer, Integer> dp = new HashMap<>();
        dp.put(0, 0);

        for (int r : rods) {
            Map<Integer, Integer> cur = new HashMap<>(dp);
            for (Map.Entry<Integer, Integer> entry : cur.entrySet()) {
                int diff = entry.getKey();
                int sum1 = entry.getValue();

                // Add rod 'r' to the first partition
                dp.put(diff + r, Math.max(dp.getOrDefault(diff + r, 0), sum1 + r));

                // Add rod 'r' to the second partition
                dp.put(diff - r, Math.max(dp.getOrDefault(diff - r, 0), sum1));
            }
        }

        return dp.getOrDefault(0, 0);
    }
}

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Problem Info

Description

Problem 956: Tallest Billboard - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:956

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a map `dp` to store the differences between the sums of the two partitions and the corresponding maximum sum of one of the subsets. Initialize `dp[0] = 0` because initially, the difference is zero and the sum is also zero.
2
Step 2: Iterate through each `rod` in the `rods` array.
3
Step 3: Create a copy `cur` of the `dp` map to avoid modifying the map while iterating over it.
4
Step 4: Iterate through the `cur` map. For each entry (difference, sum1) in `cur`:
5
Step 5: Add the current `rod` to the first partition. Update `dp[difference + rod]` with the maximum of the current value and `sum1 + rod`.
6
Step 6: Add the current `rod` to the second partition. Update `dp[difference - rod]` with the maximum of the current value and `sum1`.
7
Step 7: After processing all the rods, return `dp[0]` which contains the maximum possible height when the difference between the two partitions is 0. If `dp[0]` does not exist return 0.
8
Step 8: In the C implementation, a custom hash table is utilized to perform dynamic programming due to the lack of built in hash map functionality. It also incorporates memory management, and ensures the proper initialization of entries to differentiate them from present values.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*sum(rods))

💾

Space Complexity

O(sum(rods))

📝

Why?

The time complexity is derived from iterating through each rod (n) and for each rod, potentially updating the dp map for all possible differences (sum(rods)). The space complexity is derived from storing, at most, sum(rods) possible differences and corresponding sums in the dp map.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization can also be used to solve this problem. The recursive function would explore the choices of adding each rod to either the left or the right subset, and memoization would store the results of subproblems to avoid redundant calculations.

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

Trade-offs: While having the same time complexity, recursion can sometimes be less efficient due to function call overhead. However, it can be more intuitive for some programmers.

✨

Key Insights

1Insight 1: The problem can be modeled as finding two subsets of rods with equal sums. This is a variation of the subset sum problem.
2Insight 2: Dynamic programming is an effective approach for solving this problem. We can use a map to store the difference between the sums of the two subsets and the corresponding maximum sum of one of the subsets.
3Insight 3: The difference between the sums of the two subsets can range from -sum(rods) to +sum(rods). Thus, we need to keep track of all possible differences. The key in the dp map represents the difference.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. The solution correctly returns 0 because `dp` is initialized with {0:0}, and this value is returned when the loop is complete, as no rods were considered.
Edge case 2: All rods have length 1. The algorithm explores all combinations of adding each rod to either side. If the number of rods is even, the result will be sum(rods)/2 if number of rods is odd the result will be zero.
Edge case 3: The problem statement specifies 1 <= rods[i] <= 1000 and 1 <= rods.length <= 20, so the maximum sum of rods can be 20 * 1000 = 20000. This should be considered when choosing the size of the hash table.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the negative differences. The difference between the sums of the two subsets can be negative if the sum of the second subset is greater than the sum of the first subset.

✗Common mistake 2: Modifying the dp map while iterating over it. It's crucial to create a copy of the map before iterating and updating it to avoid unexpected behavior.

✗Common mistake 3: Not handling the case when no two equal sums can be created, leading to an incorrect result.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine allocating resources to two different projects, where each resource has a certain value. The goal is to allocate resources such that the total value allocated to each project is as close as possible.

🏦Real-world scenario 2: Load balancing. Distributing workload between two servers to achieve similar load on both sides.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Tallest Billboard"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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