What is the time complexity of Selling Pieces of Wood?

The optimal solution has a time complexity of O(m*n*(m+n)).

What is the space complexity of Selling Pieces of Wood?

The space complexity is O(m*n).

What data structures are used to solve Selling Pieces of Wood?

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

What are the key insights for solving Selling Pieces of Wood?

Insight 1: The core idea is to use dynamic programming to store the maximum profit obtainable for each possible sub-rectangle of the original wood piece. dp[i][j] represents the maximum profit for a wood piece of size i x j. Insight 2: We can build the DP table bottom-up. For each sub-rectangle, we either sell it directly if there's a price for it, or we cut it horizontally or vertically into smaller pieces and recursively calculate the maximum profit from those smaller pieces.

Which companies ask LeetCode 2312. Selling Pieces of Wood in interviews?

This problem is frequently asked at Palantir Technologies.

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Selling Pieces of Wood - LeetCode 2312 Solution

Selling Pieces of Wood - Complete Solution Guide

Selling Pieces of Wood is LeetCode problem 2312, 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 given two integers m and n that represent the height and width of a rectangular piece of wood. You are also given a 2D integer array prices , where prices[i] = [h i , w i , price i ] indicates you can sell a rectangular piece of wood of height h i and width w i for price i dollars. To cut a piece of wood, you must make a vertical or horizontal cut across the entire height or width of the piece to split it into two smaller pieces. After cutting a piece of wood into some number of smaller

Detailed Explanation

The problem asks us to find the maximum profit we can make by selling pieces of wood cut from a rectangular piece of wood with dimensions m x n. We are given a list of prices for different sized rectangular pieces. We can cut the original piece horizontally or vertically into smaller pieces. The goal is to determine the optimal cutting strategy and selling the resulting pieces to maximize profit. We cannot rotate the pieces of wood. The sizes of the pieces that can be sold and their prices are provided as input.

Solution Approach

The solution uses dynamic programming to calculate the maximum profit. We initialize a 2D array `dp` where `dp[i][j]` stores the maximum profit obtainable from a piece of wood of size `i x j`. We iterate through all possible sub-rectangles of the original wood piece and calculate the maximum profit for each. For each sub-rectangle, we consider two options: either sell it directly if its dimensions are in the `prices` list, or cut it horizontally or vertically into smaller pieces. If we cut it, we iterate through all possible cut locations and recursively calculate the maximum profit from the resulting smaller pieces. We then update `dp[i][j]` with the maximum of all these possibilities.

Step-by-Step Algorithm

Step 1: Initialize a 2D DP table `dp` of size (m+1) x (n+1) with all values set to 0. `dp[i][j]` will represent the maximum profit for a piece of wood of size `i x j`.
Step 2: Populate the DP table with the prices for the given wood sizes. Iterate through the `prices` list and set `dp[h][w] = price` for each entry `[h, w, price]`.
Step 3: Iterate through all possible wood sizes, starting from 1x1 up to m x n. For each size `h x w`, iterate through all possible horizontal cuts (from 1 to h/2) and all possible vertical cuts (from 1 to w/2).
Step 4: For each horizontal cut at `i`, the profit will be `dp[i][w] + dp[h - i][w]`. Update `dp[h][w]` with the maximum value between its current value and this profit.
Step 5: For each vertical cut at `j`, the profit will be `dp[h][j] + dp[h][w - j]`. Update `dp[h][w]` with the maximum value between its current value and this profit.
Step 6: After iterating through all possible sizes and cuts, `dp[m][n]` will contain the maximum profit obtainable from the original m x n piece of wood. Return this value.

Key Insights

Insight 1: The core idea is to use dynamic programming to store the maximum profit obtainable for each possible sub-rectangle of the original wood piece. dp[i][j] represents the maximum profit for a wood piece of size i x j.
Insight 2: We can build the DP table bottom-up. For each sub-rectangle, we either sell it directly if there's a price for it, or we cut it horizontally or vertically into smaller pieces and recursively calculate the maximum profit from those smaller pieces.
Insight 3: Consider all possible cut locations, only up to half the height or width is needed due to symmetry of choice. Cutting the rectangle i x j at height k is the same as cutting at height i-k (except for base case, when i=2k we must actually cut there to allow for more cuts later)

Complexity Analysis

Time Complexity: O(m*n*(m+n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Memoization.

Companies

Asked at: Palantir Technologies.

Selling Pieces of Wood | LeetCode 2312

Hard

Array Dynamic Programming Memoization

Time: O(m*n*(m+n))Space: O(m*n)

Solution Code

Available in:

class Solution {
    public long sellingWood(int m, int n, int[][] prices) {
        long[][] dp = new long[m + 1][n + 1];

        for (int[] price : prices) {
            int h = price[0];
            int w = price[1];
            int p = price[2];
            dp[h][w] = p;
        }

        for (int h = 1; h <= m; h++) {
            for (int w = 1; w <= n; w++) {
                for (int i = 1; i <= h / 2; i++) {
                    dp[h][w] = Math.max(dp[h][w], dp[i][w] + dp[h - i][w]);
                }

                for (int j = 1; j <= w / 2; j++) {
                    dp[h][w] = Math.max(dp[h][w], dp[h][j] + dp[h][w - j]);
                }
            }
        }

        return dp[m][n];
    }
}

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Problem Info

Description

Problem 2312: Selling Pieces of Wood - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Memoization

Companies

Palantir Technologies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2312

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

6 steps

1
Step 1: Initialize a 2D DP table `dp` of size (m+1) x (n+1) with all values set to 0. `dp[i][j]` will represent the maximum profit for a piece of wood of size `i x j`.
2
Step 2: Populate the DP table with the prices for the given wood sizes. Iterate through the `prices` list and set `dp[h][w] = price` for each entry `[h, w, price]`.
3
Step 3: Iterate through all possible wood sizes, starting from 1x1 up to m x n. For each size `h x w`, iterate through all possible horizontal cuts (from 1 to h/2) and all possible vertical cuts (from 1 to w/2).
4
Step 4: For each horizontal cut at `i`, the profit will be `dp[i][w] + dp[h - i][w]`. Update `dp[h][w]` with the maximum value between its current value and this profit.
5
Step 5: For each vertical cut at `j`, the profit will be `dp[h][j] + dp[h][w - j]`. Update `dp[h][w]` with the maximum value between its current value and this profit.
6
Step 6: After iterating through all possible sizes and cuts, `dp[m][n]` will contain the maximum profit obtainable from the original m x n piece of wood. Return this value.

💡

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(m*n*(m+n))

💾

Space Complexity

O(m*n)

📝

Why?

The nested loops for building the DP table and considering cuts define the time complexity. The space complexity directly relates to the size of the DP table needed to store 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

Memoization (Top-Down Dynamic Programming)

Instead of building the DP table bottom-up, we can use a recursive approach with memoization. We define a recursive function that takes the height and width of the wood piece as input and returns the maximum profit. We memoize the results of this function to avoid redundant calculations.

Time: O(m * n * (m + n))Space: O(m * n)

Trade-offs: Memoization can be more intuitive to implement for some, but it may have a slightly higher overhead due to function call stack. In this case, the complexities are the same, so iterative dynamic programming is generally preferred because it avoids recursion overhead.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming to store the maximum profit obtainable for each possible sub-rectangle of the original wood piece. dp[i][j] represents the maximum profit for a wood piece of size i x j.
2Insight 2: We can build the DP table bottom-up. For each sub-rectangle, we either sell it directly if there's a price for it, or we cut it horizontally or vertically into smaller pieces and recursively calculate the maximum profit from those smaller pieces.
3Insight 3: Consider all possible cut locations, only up to half the height or width is needed due to symmetry of choice. Cutting the rectangle i x j at height k is the same as cutting at height i-k (except for base case, when i=2k we must actually cut there to allow for more cuts later)

📚

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: The `prices` list is empty. In this case, the DP table will remain filled with 0s, and the function will correctly return 0, indicating no profit can be made.
Edge case 2: `m` or `n` is 1. The algorithm handles this case correctly because the loops iterate from 1 to `m` and `n` respectively. When `m` or `n` is 1, the cuts will be considered along the other dimension.
Edge case 3: A price exists for the full m x n wood, but no other prices. The initial `dp[m][n]` will be set to that price, and the cut loops won't change it, since other `dp[i][j]` are 0, ensuring we return the initial price.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize the DP table with the given prices. This will lead to incorrect profit calculations.

✗Common mistake 2: Not considering both horizontal and vertical cuts. The optimal solution might involve a combination of both types of cuts.

✗Common mistake 3: Not taking `max` while assigning `dp[h][w]` after cutting the wood will produce wrong results, as `dp[h][w]` may have been assigned by a better combination previously.

✅

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: Cutting materials like fabric, metal sheets, or glass to maximize the yield and minimize waste, where different sized pieces have different values.

🏦Real-world scenario 2: Resource allocation in manufacturing or logistics, where limited resources (e.g., warehouse space, production capacity) can be allocated to different tasks or products to maximize overall profit.

💡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: October 31, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 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 "Selling Pieces of Wood"?

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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