What is the time complexity of Maximum Height by Stacking Cuboids ?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Maximum Height by Stacking Cuboids ?

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

What are the key insights for solving Maximum Height by Stacking Cuboids ?

Insight 1: Sorting the dimensions of each cuboid is crucial. By always having the dimensions in ascending order, we standardize the representation and simplify the comparison logic later. Insight 2: Sorting the cuboids themselves based on their dimensions (e.g., width, then length, then height) allows us to efficiently determine a valid stacking order using dynamic programming. We only need to check cuboids that come before the current one in the sorted order.

Which companies ask LeetCode 1691. Maximum Height by Stacking Cuboids in interviews?

This problem is frequently asked at Samsung.

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Maximum Height by Stacking Cuboids - LeetCode 1691 Solution

Q: How to solve Maximum Height by Stacking Cuboids ?

The provided solution uses a dynamic programming approach to find the maximum stack height. First, it standardizes each cuboid by sorting its dimensions. Then, it sorts the cuboids based on their dimensions to establish a valid processing order. A dynamic programming array, `dp`, is used to store the maximum height of a stack ending with each cuboid. The solution iterates through the sorted cuboids, and for each cuboid `i`, it checks if any previous cuboid `j` can be placed beneath it. If it can, it updates `dp[i]` to be the maximum of its current value and the sum of `dp[j]` and the height of cuboid `i`. Finally, the solution returns the maximum value in the `dp` array.

Maximum Height by Stacking Cuboids - Complete Solution Guide

Maximum Height by Stacking Cuboids is LeetCode problem 1691, 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

Given n cuboids where the dimensions of the i th cuboid is cuboids[i] = [width i , length i , height i ] ( 0-indexed ). Choose a subset of cuboids and place them on each other. You can place cuboid i on cuboid j if width i <= width j and length i <= length j and height i <= height j . You can rearrange any cuboid's dimensions by rotating it to put it on another cuboid. Return the maximum height of the stacked cuboids . Example 1: Input: cuboids = [[50,45,20],[95,37,53],[45,23,12]] Output: 190 Ex

Detailed Explanation

The problem asks us to find the maximum possible height of a stack of cuboids, given a set of cuboids with varying widths, lengths, and heights. We can rearrange the dimensions of each cuboid (rotate it) to maximize its height when placed in the stack. A cuboid 'i' can be placed on top of cuboid 'j' only if the width, length, and height of 'i' are all less than or equal to the corresponding dimensions of 'j'. The goal is to select a subset of cuboids and stack them in a way that maximizes the overall height of the stack.

Solution Approach

The provided solution uses a dynamic programming approach to find the maximum stack height. First, it standardizes each cuboid by sorting its dimensions. Then, it sorts the cuboids based on their dimensions to establish a valid processing order. A dynamic programming array, `dp`, is used to store the maximum height of a stack ending with each cuboid. The solution iterates through the sorted cuboids, and for each cuboid `i`, it checks if any previous cuboid `j` can be placed beneath it. If it can, it updates `dp[i]` to be the maximum of its current value and the sum of `dp[j]` and the height of cuboid `i`. Finally, the solution returns the maximum value in the `dp` array.

Step-by-Step Algorithm

Step 1: Sort the dimensions of each cuboid to standardize the representation (width <= length <= height).
Step 2: Sort the cuboids based on their dimensions (e.g., width, then length, then height) to create a valid processing order.
Step 3: Initialize a dynamic programming array `dp` of size `n` (number of cuboids), where `dp[i]` will store the maximum height of a stack ending with cuboid `i`.
Step 4: Iterate through the sorted cuboids from `i = 0` to `n - 1`.
Step 5: For each cuboid `i`, initialize `dp[i]` to its height (cuboids[i][2]).
Step 6: Iterate through the previous cuboids from `j = 0` to `i - 1`.
Step 7: For each previous cuboid `j`, check if cuboid `i` can be placed on top of cuboid `j` (i.e., cuboids[j][0] <= cuboids[i][0] and cuboids[j][1] <= cuboids[i][1] and cuboids[j][2] <= cuboids[i][2]).
Step 8: If cuboid `i` can be placed on top of cuboid `j`, update `dp[i]` to be the maximum of its current value and `dp[j] + cuboids[i][2]`.
Step 9: After iterating through all cuboids, find the maximum value in the `dp` array, which represents the maximum height of any stack of cuboids.

Key Insights

Insight 1: Sorting the dimensions of each cuboid is crucial. By always having the dimensions in ascending order, we standardize the representation and simplify the comparison logic later.
Insight 2: Sorting the cuboids themselves based on their dimensions (e.g., width, then length, then height) allows us to efficiently determine a valid stacking order using dynamic programming. We only need to check cuboids that come before the current one in the sorted order.
Insight 3: The problem can be solved using dynamic programming, where dp[i] represents the maximum height of a stack ending with the i-th cuboid. By iterating through the sorted cuboids and checking if a previous cuboid can be placed beneath the current one, we can update dp[i] accordingly.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: Samsung.

Maximum Height by Stacking Cuboids | LeetCode 1691

Hard

Array Dynamic Programming Sorting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxHeight(int[][] cuboids) {
        int n = cuboids.length;
        
        // Standardize each cuboid by sorting its dimensions
        for (int i = 0; i < n; i++) {
            Arrays.sort(cuboids[i]);
        }
        
        // Sort all cuboids based on their dimensions.
        // This helps in the DP approach by establishing a processing order.
        Arrays.sort(cuboids, (a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else if (a[1] != b[1]) {
                return a[1] - b[1];
            } else {
                return a[2] - b[2];
            }
        });
        
        // dp[i] will store the maximum height of a stack ending with cuboid i
        int[] dp = new int[n];
        
        for (int i = 0; i < n; i++) {
            // The height of the current cuboid is its largest dimension
            dp[i] = cuboids[i][2];
            
            // Check all previous cuboids to see if they can be placed under the current one
            for (int j = 0; j < i; j++) {
                // Check if cuboid i can be placed on cuboid j
                if (cuboids[j][0] <= cuboids[i][0] &&
                    cuboids[j][1] <= cuboids[i][1] &&
                    cuboids[j][2] <= cuboids[i][2]) {
                    // If it can, we can potentially form a taller stack
                    dp[i] = Math.max(dp[i], cuboids[i][2] + dp[j]);
                }
            }
        }
        
        // The result is the maximum height found for any stack
        int max = 0;
        for (int height : dp) {
            max = Math.max(max, height);
        }
        return max;
    }
}

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Problem Info

Description

Problem 1691: Maximum Height by Stacking Cuboids - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

Samsung

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1691

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

9 steps

1
Step 1: Sort the dimensions of each cuboid to standardize the representation (width <= length <= height).
2
Step 2: Sort the cuboids based on their dimensions (e.g., width, then length, then height) to create a valid processing order.
3
Step 3: Initialize a dynamic programming array `dp` of size `n` (number of cuboids), where `dp[i]` will store the maximum height of a stack ending with cuboid `i`.
4
Step 4: Iterate through the sorted cuboids from `i = 0` to `n - 1`.
5
Step 5: For each cuboid `i`, initialize `dp[i]` to its height (cuboids[i][2]).
6
Step 6: Iterate through the previous cuboids from `j = 0` to `i - 1`.
7
Step 7: For each previous cuboid `j`, check if cuboid `i` can be placed on top of cuboid `j` (i.e., cuboids[j][0] <= cuboids[i][0] and cuboids[j][1] <= cuboids[i][1] and cuboids[j][2] <= cuboids[i][2]).
8
Step 8: If cuboid `i` can be placed on top of cuboid `j`, update `dp[i]` to be the maximum of its current value and `dp[j] + cuboids[i][2]`.
9
Step 9: After iterating through all cuboids, find the maximum value in the `dp` array, which represents the maximum height of any stack of cuboids.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates `n` times, and the inner loop iterates up to `n` times in the worst case, resulting in O(n^2). The `dp` array requires O(n) space to store the intermediate results.

📖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

Instead of using a bottom-up dynamic programming approach, we can use a top-down recursive approach with memoization. This involves defining a recursive function that calculates the maximum height of a stack ending with a given cuboid. Memoization can be used to store the results of intermediate calculations and avoid redundant computations.

Time: O(n^2)Space: O(n)

Trade-offs: Recursion with memoization can be more intuitive to implement for some developers. However, it may have a higher overhead due to function calls and recursion depth. The bottom-up DP approach may be slightly more efficient due to avoiding recursion overhead.

✨

Key Insights

1Insight 1: Sorting the dimensions of each cuboid is crucial. By always having the dimensions in ascending order, we standardize the representation and simplify the comparison logic later.
2Insight 2: Sorting the cuboids themselves based on their dimensions (e.g., width, then length, then height) allows us to efficiently determine a valid stacking order using dynamic programming. We only need to check cuboids that come before the current one in the sorted order.
3Insight 3: The problem can be solved using dynamic programming, where dp[i] represents the maximum height of a stack ending with the i-th cuboid. By iterating through the sorted cuboids and checking if a previous cuboid can be placed beneath the current one, we can update dp[i] accordingly.

📚

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.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input (cuboids is empty). The code handles this case implicitly because the loops won't execute, and the function will return 0 (or a suitable default value depending on the exact implementation details of max() function if any).
Edge case 2: All cuboids have the same dimensions. The sorting steps will still work correctly, and the dynamic programming algorithm will find the optimal stack height, which is the sum of the heights of all cuboids.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the dimensions of each cuboid. This leads to incorrect comparisons and an incorrect result.

✗Common mistake 2: Not sorting the cuboids based on their dimensions. This violates the assumption that a cuboid can only be placed on top of another cuboid if all its dimensions are less than or equal to the corresponding dimensions of the base cuboid. This leads to wrong state transitions in DP.

✗Common mistake 3: Incorrectly implementing the comparison logic for determining if one cuboid can be placed on top of another. The comparison must ensure that all three dimensions are less than or equal to each other.

✅

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: Optimizing the arrangement of boxes in a warehouse to maximize storage space, subject to constraints on box dimensions and stacking order.

🏦Real-world scenario 2: Selecting and arranging components in a system, where components have size and compatibility constraints, to maximize the performance or efficiency of the system. For example, placing servers in a rack where cooling efficiency depends on size and power consumption.

💡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: August 11, 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 13

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

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Height by Stacking Cuboids "?

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.

🤖 Need help with this problem?