Rod Cutting Problem in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum revenue vector with zeros.

DSA Rust

let mut max_revenue = vec![[1]; n + 1];

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A-1

B1

Cn

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 1 or negative values.

The maximum revenue vector should start with zeros because initially, no revenue is earned.

2fill in blank

medium

Complete the code to iterate over all rod lengths from 1 to n.

DSA Rust

for length in 1..=[1] {

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An

Blength

Cprices.len()

Dmax_revenue.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using prices length or max_revenue length incorrectly.

We iterate from 1 to n inclusive to calculate maximum revenue for each rod length.

3fill in blank

hard

Fix the error in the code to update the maximum revenue for the current length.

DSA Rust

max_revenue[length] = max_revenue[length].max(prices[i - 1] + max_revenue[[1]]);

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Alength - i

Blength

Ci

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i or length directly instead of length - i.

The remaining rod length after cutting is length - i, so we use that to get the max revenue for the leftover piece.

4fill in blank

hard

Fill both blanks to complete the inner loop iterating over possible first cut lengths.

DSA Rust

for [1] in 1..=[2] {

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Ai

Blength

Cj

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names or loop limits.

The inner loop variable i represents the first cut length, iterating from 1 to the current rod length.

5fill in blank

hard

Fill all three blanks to complete the function signature and return statement.

DSA Rust

fn rod_cutting(prices: &[i32], [1]: usize) -> [2] {
    // implementation
    [3]
}

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An

Bi32

Cmax_revenue[n]

Dusize

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong parameter types or return values.

The function takes the rod length n as usize, returns i32 revenue, and returns max_revenue[n] at the end.

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