Rod Cutting Problem in DSA Rust - Build from Scratch

Rod Cutting Problem

📖 Scenario: Imagine you have a wooden 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 highest total price.

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

📋 What You'll Learn

Create a vector called prices with prices for rod lengths 1 to 8

Create a variable called rod_length with the value 8

Use a vector called max_prices to store maximum prices for each length

Use a for loop with variable length from 1 to rod_length

Use a nested for loop with variable cut from 1 to length

Calculate the maximum price for each length by checking all cuts

Print the maximum price for the full rod length

💡 Why This Matters

🌍 Real World

This problem models how to maximize profit when selling materials that can be cut into smaller pieces, like wood, metal rods, or fabric.

💼 Career

Understanding this problem helps in learning dynamic programming, which is useful in software engineering roles involving optimization and algorithm design.

1

Create the prices vector

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

DSA Rust

# Create the prices vector
# Your code here

Hint

Use let prices = vec![...] to create the vector with the exact prices.

2

Set the rod length

Create a variable called rod_length and set it to 8

DSA Rust

fn main() {
    let prices = vec![1, 5, 8, 9, 10, 17, 17, 20];
    // Set the rod length
    // Your code here
}

Hint

Use let rod_length = 8; to set the rod length.

3

Calculate maximum prices for each length

Create a mutable vector called max_prices with length rod_length + 1 filled with zeros. Use a for loop with variable length from 1 to rod_length inclusive. Inside it, use a nested for loop with variable cut from 1 to length inclusive. Update max_prices[length] with the maximum of its current value and prices[cut - 1] + max_prices[length - cut].

DSA Rust

fn main() {
    let prices = vec![1, 5, 8, 9, 10, 17, 17, 20];
    let rod_length = 8;
    // Create max_prices vector
    // Use nested loops to calculate max prices
    // Your code here
}

Hint

Use two for loops and update max_prices[length] with the best price found.

4

Print the maximum price for the full rod length

Print the maximum price for the full rod length using println!("{}", max_prices[rod_length])

DSA Rust

fn main() {
    let prices = vec![1, 5, 8, 9, 10, 17, 17, 20];
    let rod_length = 8;
    let mut max_prices = vec![0; rod_length + 1];
    for length in 1..=rod_length {
        let mut max_val = 0;
        for cut in 1..=length {
            let val = prices[cut - 1] + max_prices[length - cut];
            if val > max_val {
                max_val = val;
            }
        }
        max_prices[length] = max_val;
    }
    // Print the maximum price
    // Your code here
}

Hint

Use println!("{}", max_prices[rod_length]) to show the result.

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:

   Goal = maximize profit [OK]

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

Common Mistakes:

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

2. Which of the following Rust function signatures correctly represents a function to solve the rod cutting problem using dynamic programming?

easy

A. fn rod_cutting(prices: &[i32], n: usize) -> i32

B. fn rod_cutting(prices: Vec<i32>, n: i32) -> usize

C. fn rod_cutting(prices: &mut [i32], n: usize) -> usize

D. fn rod_cutting(prices: &[i32], n: i32) -> Vec<i32>

Solution

1. Step 1: Analyze parameter types

   The prices array should be a slice reference &[i32] for read-only access, and the rod length n should be usize for indexing.

2. Step 2: Analyze return type

   The function returns the maximum profit as an i32 integer.

3. Final Answer:

   fn rod_cutting(prices: &[i32], n: usize) -> i32 -> Option A

4. Quick Check:

   Correct signature uses &[i32], usize, returns i32 [OK]

Hint: Use slice &[i32] and usize for length, return i32 profit [OK]

Common Mistakes:

• Using Vec instead of slice for prices
• Using i32 for length parameter
• Returning Vec instead of profit integer

3. What is the output of the following Rust code snippet for rod cutting?

fn rod_cutting(prices: &[i32], n: usize) -> i32 {
    let mut dp = vec![0; n + 1];
    for i in 1..=n {
        let mut max_val = i32::MIN;
        for j in 1..=i {
            max_val = max_val.max(prices[j - 1] + dp[i - j]);
        }
        dp[i] = max_val;
    }
    dp[n]
}

fn main() {
    let prices = [1, 5, 8, 9];
    let n = 4;
    println!("{}", rod_cutting(&prices, n));
}

medium

A. 8

B. 9

C. 5

D. 10

Solution

1. Step 1: Understand the DP array construction

   For n=4, dp array stores max profit for lengths 0 to 4. Iterations compute max of cutting at all positions.

2. Step 2: Calculate dp values step-by-step

   dp[1] = max(1) = 1 dp[2] = max(1+1,5) = 5 dp[3] = max(1+5,5+1,8) = 8 dp[4] = max(1+8,5+5,8+1,9) = 10

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Calculate dp array stepwise for each rod length [OK]

Common Mistakes:

• Using wrong indices for prices
• Not considering all cut positions
• Confusing max and min values

4. Identify the bug in this Rust function for rod cutting:

fn rod_cutting(prices: &[i32], n: usize) -> i32 {
    let mut dp = vec![0; n];
    for i in 1..=n {
        let mut max_val = i32::MIN;
        for j in 1..=i {
            max_val = max_val.max(prices[j - 1] + dp[i - j]);
        }
        dp[i] = max_val;
    }
    dp[n]
}

medium

A. prices array should be mutable

B. dp vector size should be n+1 to avoid index out of bounds

C. Loop ranges should start from 0 instead of 1

D. Return type should be usize instead of i32

Solution

1. Step 1: Check dp vector initialization

   dp is initialized with size n, but code accesses dp[i] where i can be n, causing out of bounds error.

2. Step 2: Correct dp size

   dp should be size n+1 to safely store dp[n] without error.

3. Final Answer:

   dp vector size should be n+1 to avoid index out of bounds -> Option B

4. Quick Check:

   dp size = n+1 fixes index error [OK]

Hint: dp array size must be n+1 for indexing up to n [OK]

Common Mistakes:

• Ignoring dp size causing runtime panic
• Changing prices to mutable unnecessarily
• Incorrect loop ranges causing logic 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 using the rod cutting problem approach?

hard

A. Maximum profit is 13 with cuts [1, 1, 3]

B. Maximum profit is 11 with cuts [1, 4]

C. Maximum profit is 12 with cuts [2, 3]

D. Maximum profit is 10 with no cuts

Solution

1. Step 1: Calculate max profit dp array

   dp[1]=2, dp[2]=max(2+2,5)=5, dp[3]=max(2+5,5+2,7)=7, dp[4]=max(2+7,5+5,7+2,8)=10, dp[5]=max(2+10,5+7,7+5,8+2,10)=12

2. Step 2: Determine cuts for max profit 12

   Cutting rod into lengths 2 and 3 gives prices 5 + 7 = 12, which is maximum.

3. Final Answer:

   Maximum profit is 12 with cuts [2, 3] -> Option C

4. Quick Check:

   Profit 12 from cuts 2 and 3 [OK]

Hint: Check all cut combinations to find max profit and cuts [OK]

Common Mistakes:

• Choosing no cuts for max profit
• Miscomputing dp values
• Ignoring cut combinations for max profit