Rod Cutting Problem in DSA Rust - Execution Trace

Concept Flow - Rod Cutting Problem

Start with full rod length

↓

Try all cut lengths from 1 to n

↓

Calculate price for current cut + best price for remaining rod

↓

Compare with max price so far

↓

Update max price if current is better

↓

Repeat for all cut lengths

↓

Store max price for current rod length

↓

Return max price for full rod

We try all possible first cuts, calculate total price by adding cut price and best price for remaining rod, keep track of max price, and store results to avoid recomputation.

Execution Sample

DSA Rust

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

This code finds the maximum price obtainable by cutting a rod of length n using dynamic programming.

Execution Table

Step	Operation	Current Rod Length	Cut Length Tried	Price for Cut	Best Price for Remaining	Max Price So Far	dp Array State	Visual State
1	Initialize dp array	0	-	-	-	0	[0,0,0,0,0]	dp: [0,0,0,0,0]
2	Length=1, cut=1	1	1	prices[0]=2	dp[0]=0	2	[0,2,0,0,0]	dp: [0,2,0,0,0]
3	Length=2, cut=1	2	1	prices[0]=2	dp[1]=2	4	[0,2,4,0,0]	dp: [0,2,4,0,0]
4	Length=2, cut=2	2	2	prices[1]=5	dp[0]=0	5	[0,2,5,0,0]	dp: [0,2,5,0,0]
5	Length=3, cut=1	3	1	prices[0]=2	dp[2]=5	7	[0,2,5,7,0]	dp: [0,2,5,7,0]
6	Length=3, cut=2	3	2	prices[1]=5	dp[1]=2	7	[0,2,5,7,0]	dp: [0,2,5,7,0]
7	Length=3, cut=3	3	3	prices[2]=7	dp[0]=0	7	[0,2,5,7,0]	dp: [0,2,5,7,0]
8	Length=4, cut=1	4	1	prices[0]=2	dp[3]=7	9	[0,2,5,7,9]	dp: [0,2,5,7,9]
9	Length=4, cut=2	4	2	prices[1]=5	dp[2]=5	10	[0,2,5,7,10]	dp: [0,2,5,7,10]
10	Length=4, cut=3	4	3	prices[2]=7	dp[1]=2	10	[0,2,5,7,10]	dp: [0,2,5,7,10]
11	Length=4, cut=4	4	4	prices[3]=8	dp[0]=0	10	[0,2,5,7,10]	dp: [0,2,5,7,10]
12	Return max price for length 4	4	-	-	-	10	[0,2,5,7,10]	Max price: 10

💡 All rod lengths from 1 to 4 processed, dp array fully computed.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 7	After Step 11	Final
dp	[0,0,0,0,0]	[0,2,0,0,0]	[0,2,5,0,0]	[0,2,5,7,0]	[0,2,5,7,10]	[0,2,5,7,10]
max_val	-	2	5	7	10	10
length	-	1	2	3	4	4
cut	-	1	2	3	4	-

Key Moments - 3 Insights

Why do we try all cut lengths from 1 to current rod length?

Because each cut length represents a possible first cut, and we want to find the maximum price by considering all options. See execution_table rows 3-11 where each cut length is tried.

Why do we add prices[cut - 1] and dp[length - cut]?

prices[cut - 1] is the price for the first cut piece, dp[length - cut] is the best price for the remaining rod. Adding them gives total price for that cut. This is shown in execution_table columns 'Price for Cut' and 'Best Price for Remaining'.

Why do we store max price in dp[length]?

To remember the best price for a rod of that length so we don't recompute it again. This is the core of dynamic programming, shown in dp array updates in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the max price so far for rod length 3 after cut length 1?

A 4

B 7

C 6

D 2

💡 Hint

Check the 'Max Price So Far' column at step 5 in execution_table.

At which step does the dp array first contain the value 7?

A Step 4

B Step 7

C Step 5

D Step 9

💡 Hint

Look at the 'dp Array State' column and find when 7 appears first.

If the price for cut length 4 was increased to 10, how would the max price for rod length 4 change in the dp array?

A It would remain 10

B It would become 10

C It would become 12

D It would become 8

💡 Hint

Check step 11 where cut length 4 price is compared to max price so far.

Concept Snapshot

Rod Cutting Problem:
- Goal: Maximize price by cutting rod into pieces
- Try all cut lengths from 1 to n
- Use dp array to store max price for each length
- dp[length] = max(prices[cut-1] + dp[length-cut])
- Bottom-up dynamic programming approach
- Time complexity: O(n^2)

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We use a dp array where dp[length] stores the best price for that length. For each rod length from 1 to n, we try all possible first cuts. For each cut, we add the price of that cut piece to the best price of the remaining rod length. We keep track of the maximum price found and store it in dp[length]. After filling dp for all lengths, dp[n] gives the maximum price for the full rod. This approach avoids repeated calculations by storing intermediate results. The execution table shows each step, the cuts tried, prices calculated, and how dp array updates. Key moments clarify why all cuts are tried, why prices are added, and why dp stores max prices. The visual quiz tests understanding of dp updates and max price calculations. This method runs in O(n^2) time and is a classic example of dynamic programming.

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