How do you solve Selling Pieces of Wood on LeetCode?

Selling Pieces of Wood (LeetCode #2312) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * (m + n)) time complexity and O(m * n) space complexity. Key insight: Recursive cutting leads to overlapping subproblems.

What is the brute force solution for Selling Pieces of Wood?

The brute force approach for Selling Pieces of Wood is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible way to cut the wood and calculate the maximum profit we can earn. This involves recursively checking all possible cuts and combinations of selling pieces. The optimal Optimal Solution approach improves this to O(m * n * (m + n)).

What algorithm or data structure is used to solve Selling Pieces of Wood?

Selling Pieces of Wood uses the following concepts: Array, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Memoization.

What are the interview tips for Selling Pieces of Wood?

Explain your thought process clearly. Discuss the trade-offs of each approach. Practice writing clean and efficient code.

What are common mistakes when solving Selling Pieces of Wood?

Not considering all possible cuts. Forgetting to check direct selling prices.

#2312

Selling Pieces of Wood

Hard

ArrayDynamic ProgrammingMemoizationDynamic ProgrammingMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n * (m + n))
Space	O(1)	O(m * n)

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Intuition

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

We can use dynamic programming to store the maximum profit for each piece of wood size, avoiding redundant calculations. This allows us to build up the solution efficiently.

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Algorithm

4 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the maximum profit for a piece of wood of size i x j.
2Step 2: Initialize dp with direct sell prices from the prices array.
3Step 3: Iterate through all possible heights and widths, updating dp[i][j] based on possible cuts.
4Step 4: For each cut, calculate the profit from the two resulting pieces and update dp[i][j] with the maximum profit.

solution.py12 lines

1def maxProfit(m, n, prices):
2    dp = [[0] * (n + 1) for _ in range(m + 1)]
3    price_map = {(h, w): price for h, w, price in prices}
4    for (h, w), price in price_map.items():
5        dp[h][w] = max(dp[h][w], price)
6    for height in range(1, m + 1):
7        for width in range(1, n + 1):
8            for h in range(1, height):
9                dp[height][width] = max(dp[height][width], dp[h][width] + dp[height - h][width])
10            for w in range(1, width):
11                dp[height][width] = max(dp[height][width], dp[height][w] + dp[height][width - w])
12    return dp[m][n]

ℹ

Complexity note: The time complexity is O(m * n * (m + n)) because we iterate through all dimensions and for each dimension, we check all possible cuts, leading to a cubic-like growth in operations.

1Recursive cutting leads to overlapping subproblems.
2Dynamic programming helps store results to avoid redundant calculations.

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