Rod Cutting Problem in DSA Python - Build from Scratch

Rod Cutting Problem

📖 Scenario: You have a metal rod of a certain length. You want to cut it into smaller pieces to sell. Each piece has a price depending on its length. Your goal is to find the best way to cut the rod to get the most money.

🎯 Goal: Build a program that calculates the maximum profit you can get by cutting the rod into pieces using given prices for each length.

📋 What You'll Learn

Create a list called prices with prices for rod lengths 1 to 8.

Create a variable called rod_length with the value 8.

Use a list called max_profit to store the best profit for each length from 0 to rod_length.

Use a loop to calculate the maximum profit for each rod length from 1 to rod_length.

Print the maximum profit for the full rod length.

💡 Why This Matters

🌍 Real World

This problem models how to cut raw materials like metal rods or wood to maximize sales profit.

💼 Career

Understanding this problem helps in optimization tasks and dynamic programming, useful in software engineering and operations research.

1

Create the price list

Create a list called prices with these exact values: [1, 5, 8, 9, 10, 17, 17, 20].

DSA Python

# Create the prices list for rod lengths 1 to 8
# Your code here

Hint

The list prices holds the price for each rod length from 1 to 8.

2

Set the rod length

Create a variable called rod_length and set it to 8.

DSA Python

prices = [1, 5, 8, 9, 10, 17, 17, 20]
# Set the rod length to 8
# Your code here

Hint

The variable rod_length tells us how long the rod is.

3

Calculate maximum profit for each length

Create a list called max_profit with zeros of length rod_length + 1. Then use a for loop with variable i from 1 to rod_length inclusive. Inside it, use another for loop with variable j from 0 to i-1 inclusive. Update max_profit[i] to the maximum of its current value and prices[j] + max_profit[i - j - 1].

DSA Python

prices = [1, 5, 8, 9, 10, 17, 17, 20]
rod_length = 8
# Calculate max profit for each rod length
# Your code here

Hint

Use two loops: outer loop for rod length i, inner loop for cut position j. Update max_profit[i] with the best price.

4

Print the maximum profit

Print the value of max_profit[rod_length] to show the maximum profit for the full rod length.

DSA Python

prices = [1, 5, 8, 9, 10, 17, 17, 20]
rod_length = 8
max_profit = [0] * (rod_length + 1)
for i in range(1, rod_length + 1):
    for j in range(i):
        max_profit[i] = max(max_profit[i], prices[j] + max_profit[i - j - 1])
# Print the maximum profit for the full rod length
# Your code here

Hint

The maximum profit for a rod of length 8 is 22.

Practice

1. What is the main goal of the Rod Cutting Problem in algorithms?

easy

A. To maximize the total profit by cutting the rod into pieces

B. To minimize the number of cuts made on the rod

C. To find the shortest rod length possible

D. To count the total number of ways to cut the rod

Solution

1. Step 1: Understand the problem objective

   The rod cutting problem aims to cut a rod into pieces to get the maximum total price from selling those pieces.

2. Step 2: Identify the goal

   The goal is to maximize profit, not minimize cuts or count ways.

3. Final Answer:

   To maximize the total profit by cutting the rod into pieces -> Option A

4. Quick Check:

   Maximize profit = maximize total price [OK]

Hint: Focus on maximizing profit, not minimizing cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing the problem with counting ways
• Assuming shortest rod length is the goal

2. Which of the following is the correct way to initialize the DP array for rod cutting of length n in Python?

easy

A. dp = [0] * (n + 1)

B. dp = [0] * n

C. dp = [null] * n

D. dp = []

Solution

1. Step 1: Understand DP array size

   We need to store results for rod lengths from 0 to n, so array size must be n+1.

2. Step 2: Choose correct initialization

   Initializing with zeros for all lengths including 0 is done by dp = [0] * (n + 1).

3. Final Answer:

   dp = [0] * (n + 1) -> Option A

4. Quick Check:

   DP array size = n+1 with zeros [OK]

Hint: DP array size must cover 0 to n lengths [OK]

Common Mistakes:

• Using dp size n instead of n+1
• Initializing with null instead of 0
• Leaving dp empty without size

3. What will be the output of the following code snippet for rod length 4 and prices [1, 5, 8, 9]?

def rod_cutting(prices, n):
    dp = [0] * (n + 1)
    for i in range(1, n + 1):
        max_val = -1
        for j in range(i):
            max_val = max(max_val, prices[j] + dp[i - j - 1])
        dp[i] = max_val
    return dp[n]

print(rod_cutting([1, 5, 8, 9], 4))

medium

A. 9

B. 10

C. 8

D. 5

Solution

1. Step 1: Trace DP array calculation

   For n=4, prices=[1,5,8,9], calculate dp values: - dp[1] = max(1 + dp[0]) = 1 - dp[2] = max(1 + dp[1], 5 + dp[0]) = max(1+1,5+0)=5 - dp[3] = max(1 + dp[2], 5 + dp[1], 8 + dp[0]) = max(1+5,5+1,8+0)=8 - dp[4] = max(1 + dp[3], 5 + dp[2], 8 + dp[1], 9 + dp[0]) = max(1+8,5+5,8+1,9+0)=max(9,10,9,9)=10

2. Step 2: Return final dp value

   The function returns dp[4] = 10, which is the maximum profit.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Check all cuts and pick max sum for each length [OK]

Common Mistakes:

• Ignoring smaller cuts and taking only full length price
• Off-by-one errors in indexing prices or dp
• Returning dp[n-1] instead of dp[n]

4. Identify the error in the following rod cutting code snippet:

def rod_cut(prices, n):
    dp = [0] * n
    for i in range(1, n + 1):
        max_val = 0
        for j in range(i):
            max_val = max(max_val, prices[j] + dp[i - j - 1])
        dp[i] = max_val
    return dp[n]

print(rod_cut([2, 5, 7, 8], 4))

medium

A. max_val should start from -1

B. Loop range should start from 0, not 1

C. dp array size is incorrect, should be n+1

D. prices indexing is off by one

Solution

1. Step 1: Check dp array initialization

   dp is initialized with size n, but dp[i] accesses index n, causing IndexError.

2. Step 2: Correct dp size

   dp should be size n+1 to store results for lengths 0 to n.

3. Final Answer:

   dp array size is incorrect, should be n+1 -> Option C

4. Quick Check:

   dp size = n+1 to avoid index errors [OK]

Hint: DP array must cover all lengths including zero [OK]

Common Mistakes:

• Using dp size n instead of n+1
• Starting max_val from 0 instead of -1 (not critical here)
• Incorrect loop ranges causing index errors

5. Given a rod of length 5 and prices [2, 5, 7, 8, 10], what is the maximum profit and the cuts to achieve it? Choose the correct maximum profit and one valid cut combination.

hard

A. Maximum profit: 10; Cuts: [5]

B. Maximum profit: 14; Cuts: [1, 4]

C. Maximum profit: 13; Cuts: [3, 2]

D. Maximum profit: 12; Cuts: [2, 3]

Solution

1. Step 1: Calculate max profit using DP

   Prices: length 1=2, 2=5, 3=7, 4=8, 5=10 Calculate dp: - dp[1]=2 - dp[2]=max(2+dp[1],5)=max(2+2,5)=5 - dp[3]=max(2+dp[2],5+dp[1],7)=max(2+5,5+2,7)=7 - dp[4]=max(2+dp[3],5+dp[2],7+dp[1],8)=max(2+7,5+5,7+2,8)=10 - dp[5]=max(2+dp[4],5+dp[3],7+dp[2],8+dp[1],10)=max(2+10,5+7,7+5,8+2,10)=max(12,12,12,10,10)=12

2. Step 2: Identify cuts for max profit 12

   One valid cut is [2, 3] because price[2]=5 and price[3]=7, total 12.

3. Step 3: Check if better cuts exist

   Check cut [1,4]: price[1]=2 + price[4]=8 = 10, less than 12. Check cut [3,2]: price[3]=7 + price[2]=5 = 12. Check cut [5]: price[5]=10. No combination yields 14.

4. Final Answer:

   Maximum profit: 12; Cuts: [2, 3] -> Option D

5. Quick Check:

   Max profit 12 from cuts 2 and 3 [OK]

Hint: Try all cut pairs to find max sum [OK]

Common Mistakes:

• Assuming full length price is always max
• Ignoring multiple cut combinations
• Mixing up prices and lengths