Rod Cutting Problem in DSA Rust

Mental Model

We want to cut a rod into pieces to get the most money by selling those pieces. Each piece length has a price, and we try all ways to cut to find the best total price.

Analogy: Imagine you have a chocolate bar and different prices for each piece size. You want to break it into parts to sell and get the most money, like selling candy pieces at a fair.

Rod length: 8 units
Prices: [1, 5, 8, 9, 10, 17, 17, 20]

Initial rod:
[1][2][3][4][5][6][7][8]
↑ full rod length 8

Dry Run Walkthrough

Input: rod length 4, prices for lengths 1 to 4: [1, 5, 8, 9]

Goal: Find the best way to cut the rod of length 4 to get maximum price

Step 1: Try no cut: sell whole rod length 4 for price 9

Max price so far = 9 for length 4

Why: Start with the option of selling the rod as is

Step 2: Cut rod into lengths 1 and 3; price = price(1) + max price(3) = 1 + ?

Need max price for length 3 to calculate

Why: Try cutting rod into two parts to see if better price

Step 3: Calculate max price for length 3: try no cut price(3) = 8

Max price for length 3 = 8

Why: Find max price for smaller rod to use in bigger rod calculation

Step 4: Calculate price for cut lengths 1 and 3: 1 + 8 = 9

Price for cut (1,3) = 9 equals current max 9

Why: Compare cut price with current max to update

Step 5: Cut rod into lengths 2 and 2; price = price(2) + max price(2) = 5 + ?

Need max price for length 2

Why: Try another cut option

Step 6: Calculate max price for length 2: try no cut price(2) = 5

Max price for length 2 = 5

Why: Find max price for smaller rod

Step 7: Calculate price for cut (2,2): 5 + 5 = 10

Price for cut (2,2) = 10 is better than current max 9

Why: Update max price for length 4 to 10

Step 8: Cut rod into lengths 3 and 1; price = price(3) + max price(1) = 8 + 1 = 9

Price for cut (3,1) = 9 less than max 10

Why: Check last cut option

Result:

Max price for rod length 4 = 10
Best cut: two pieces of length 2
[2] -> [2] -> null

Annotated Code

DSA Rust

struct RodCutting {
    prices: Vec<i32>,
}

impl RodCutting {
    fn new(prices: Vec<i32>) -> Self {
        Self { prices }
    }

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

fn main() {
    let prices = vec![1, 5, 8, 9];
    let rod_length = 4;
    let rod_cutting = RodCutting::new(prices);
    let max_price = rod_cutting.max_profit(rod_length);
    println!("Max price for rod length {} = {}", rod_length, max_price);
}

for i in 1..=length {

iterate over rod lengths from 1 to target length

for j in 1..=i {

try all first cut lengths j for current rod length i

let current = self.prices[j - 1] + dp[i - j];

calculate price by cutting first piece j and adding best price for remaining length

if current > max_val { max_val = current; }

update max price for rod length i if current cut is better

dp[i] = max_val;

store max price for rod length i

Output

Max price for rod length 4 = 10

Complexity Analysis

Time: O(n^2) because for each rod length we try all possible first cuts

Space: O(n) for dp array storing max prices for all lengths up to n

vs Alternative: Naive recursive approach is exponential due to repeated calculations; DP reduces it to polynomial time by storing results

Edge Cases

rod length 0

returns 0 because no rod means no profit

DSA Rust

dp = vec![0; length + 1];

rod length 1

returns price of length 1 piece directly

DSA Rust

for i in 1..=length { ... }

prices vector shorter than rod length

code assumes prices for all lengths up to rod length; if shorter, may panic or incorrect result

DSA Rust

None (input must have prices for all lengths)

When to Use This Pattern

When you see a problem asking to maximize profit by cutting or splitting a resource into parts with given prices, reach for the Rod Cutting DP pattern because it tries all cuts and remembers best results.

Common Mistakes

Mistake: Not storing intermediate results and recalculating max profit for same lengths repeatedly
Fix: Use a dp array to store max profit for each length to avoid repeated work

Mistake: Using 0-based indexing incorrectly for prices array
Fix: Adjust indices carefully: price for length j is prices[j-1]

Summary

Finds the best way to cut a rod to maximize total selling price.

Use when you have prices for pieces and want to maximize profit by cutting.

Try all cuts and remember best results for smaller lengths to build up solution.

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