Rod Cutting Problem in DSA Go - Execution Trace

Concept Flow - Rod Cutting Problem

Start with full rod length

↓

Try all cut lengths i from 1 to n

↓

Calculate price for cut i + best price for remaining length

↓

Choose max price among all cuts

↓

Store max price in DP table for current length

↓

Repeat for all lengths up to n

↓

Result: max price for full rod length

We try all possible first cuts, calculate prices recursively, store best results in a table, and build up to the full rod length.

Execution Sample

DSA Go

prices := []int{0,1,5,8,9}
length := 4
maxPrice := rodCut(prices, length)

func rodCut(p []int, n int) int {
  dp := make([]int, n+1)
  for i := 1; i <= n; i++ {
    maxVal := 0
    for j := 1; j <= i; j++ {
      val := p[j] + dp[i-j]
      if val > maxVal {
        maxVal = val
      }
    }
    dp[i] = maxVal
  }
  return dp[n]
}

This code finds the best price to cut a rod of length 4 using given prices for lengths 1 to 4.

Execution Table

Step	Operation	Current Length i	Cut Length j	Price p[j]	dp[i-j]	Calculated Value	Max Value So Far	dp Array State	Visual State
1	Initialize dp	-	-	-	-	-	0	[0,0,0,0,0]	dp initialized with zeros
2	i=1, j=1	1	1	1	0	1	1	[0,1,0,0,0]	dp[1] updated to 1
3	i=2, j=1	2	1	1	1	2	2	[0,1,2,0,0]	dp[2] updated to 2
4	i=2, j=2	2	2	5	0	5	5	[0,1,5,0,0]	dp[2] updated to 5 (max)
5	i=3, j=1	3	1	1	5	6	6	[0,1,5,6,0]	dp[3] updated to 6
6	i=3, j=2	3	2	5	1	6	6	[0,1,5,6,0]	dp[3] remains 6
7	i=3, j=3	3	3	8	0	8	8	[0,1,5,8,0]	dp[3] updated to 8 (max)
8	i=4, j=1	4	1	1	8	9	9	[0,1,5,8,9]	dp[4] updated to 9
9	i=4, j=2	4	2	5	5	10	10	[0,1,5,8,10]	dp[4] updated to 10 (max)
10	i=4, j=3	4	3	8	1	9	10	[0,1,5,8,10]	dp[4] remains 10
11	i=4, j=4	4	4	9	0	9	10	[0,1,5,8,10]	dp[4] remains 10
12	Return result	4	-	-	-	-	-	[0,1,5,8,10]	Maximum price for length 4 is 10

💡 All 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 9	Final
i	-	1	2	3	4	4
j	-	1	2	3	2	4
maxVal	0	1	5	8	10	10
dp	[0,0,0,0,0]	[0,1,0,0,0]	[0,1,5,0,0]	[0,1,5,8,0]	[0,1,5,8,10]	[0,1,5,8,10]

Key Moments - 3 Insights

Why do we check all cut lengths j from 1 to i for each rod length i?

Because the best price for length i might come from cutting the rod into different pieces. Checking all j ensures we find the maximum price by considering every possible first cut, as shown in execution_table rows 3 to 11.

Why do we add p[j] + dp[i-j] instead of just p[i]?

p[j] is the price for the first cut of length j, and dp[i-j] is the best price for the remaining rod length. Adding them combines the price of the first piece and the best price for the rest, building the solution step-by-step (see execution_table rows 3-11).

Why do we store results in dp array instead of recalculating?

Storing results in dp avoids repeated calculations for the same lengths, making the solution efficient. This is why dp array values grow and are reused, as seen in variable_tracker and execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the dp array state after step 7?

A [0,1,5,8,10]

B [0,1,5,6,0]

C [0,1,5,8,0]

D [0,1,2,0,0]

💡 Hint

Check the 'dp Array State' column in row 7 of execution_table.

At which step does dp[4] first get updated to 10?

A Step 8

B Step 9

C Step 10

D Step 11

💡 Hint

Look at 'Max Value So Far' and 'dp Array State' columns for i=4 in execution_table rows 8-11.

If price for length 2 (p[2]) was 6 instead of 5, how would dp[2] change after step 4?

A dp[2] would be 6

B dp[2] would remain 5

C dp[2] would be 1

D dp[2] would be 0

💡 Hint

Check calculation val = p[j] + dp[i-j] for i=2, j=2 in execution_table row 4.

Concept Snapshot

Rod Cutting Problem:
- Given prices for rod lengths, find max revenue by cutting.
- Use dp array to store max price for each length.
- For each length i, try all cuts j (1 to i).
- dp[i] = max(dp[i], price[j] + dp[i-j]).
- Result is dp[n] for rod length n.

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We use a dp array to store the best prices for smaller lengths. For each rod length i, we try all possible first cuts j and calculate price[j] plus the best price for the remaining length dp[i-j]. We keep the maximum value found and store it in dp[i]. After processing all lengths up to n, dp[n] holds the maximum price for the full rod. This approach avoids repeated calculations by storing intermediate results. The execution table shows each step updating dp and choosing the best cuts. Key moments clarify why all cuts are tried, why prices are added, and why dp storage is important. The visual quiz tests understanding of dp states and updates.

Practice

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

easy

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

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

C. To find the shortest rod length possible

D. To minimize the number of cuts made on 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 B

4. Quick Check:

   Maximize profit = To maximize the total profit by cutting the rod into pieces [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 Go function signatures correctly represents a rod cutting function that takes rod length and price slice as input and returns maximum profit as int?

easy

A. func rodCutting(prices int, length []int) int

B. func rodCutting(prices []int, length int) bool

C. func rodCutting(length int, prices []int) string

D. func rodCutting(length int, prices []int) int

Solution

1. Step 1: Analyze parameter types and order

   The rod length should be an int, prices should be a slice of ints ([]int), and return type should be int for max profit.

2. Step 2: Match correct signature

   func rodCutting(length int, prices []int) int matches: func rodCutting(length int, prices []int) int

3. Final Answer:

   func rodCutting(length int, prices []int) int -> Option D

4. Quick Check:

   Correct parameter types and return int = func rodCutting(length int, prices []int) int [OK]

Hint: Remember: length int, prices []int, return int for profit [OK]

Common Mistakes:

• Swapping parameter order
• Wrong return type like string or bool
• Using incorrect parameter types

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

package main

import "fmt"

func rodCutting(n int, prices []int) int {
    dp := make([]int, n+1)
    for i := 1; i <= n; i++ {
        maxVal := 0
        for j := 1; j <= i; j++ {
            val := prices[j-1] + dp[i-j]
            if val > maxVal {
                maxVal = val
            }
        }
        dp[i] = maxVal
    }
    return dp[n]
}

func main() {
    prices := []int{1, 5, 8, 9}
    result := rodCutting(4, prices)
    fmt.Println(result)
}

medium

A. 8

B. 9

C. 10

D. 11

Solution

1. Step 1: Understand the dp array calculation

   dp[i] stores max profit for rod length i. For length 4, check all cuts: - Cut 1: price 1 + dp[3] = 1 + 8 = 9 - Cut 2: price 5 + dp[2] = 5 + 5 = 10 - Cut 3: price 8 + dp[1] = 8 + 1 = 9 - Cut 4: price 9 + dp[0] = 9 + 0 = 9

2. Step 2: Calculate dp values stepwise

   dp[1]=1, dp[2]=5, dp[3]=8, dp[4]=max(9,10,9,9)=10 But re-check dp[3]: - Cut 1: 1 + dp[2]=1+5=6 - Cut 2: 5 + dp[1]=5+1=6 - Cut 3: 8 + dp[0]=8+0=8 So dp[3]=8 Now dp[4]: - Cut 1: 1 + dp[3]=1+8=9 - Cut 2: 5 + dp[2]=5+5=10 - Cut 3: 8 + dp[1]=8+1=9 - Cut 4: 9 + dp[0]=9+0=9 Max is 10

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Calculate dp array stepwise checking all cuts [OK]

Common Mistakes:

• Ignoring dp values for smaller lengths
• Choosing wrong max profit among cuts
• Confusing indices in prices slice

4. Identify the bug in the following Go code for rod cutting and select the correct fix:

func rodCutting(n int, prices []int) int {
    dp := make([]int, n)
    for i := 1; i <= n; i++ {
        maxVal := 0
        for j := 1; j <= i; j++ {
            val := prices[j-1] + dp[i-j]
            if val > maxVal {
                maxVal = val
            }
        }
        dp[i] = maxVal
    }
    return dp[n]
}

medium

A. Change dp size to n+1 and access dp[i] safely

B. Change loop to start from i := 0 instead of 1

C. Remove inner loop and use only prices[i-1]

D. Change return to dp[n-1] instead of dp[n]

Solution

1. Step 1: Identify index out of range error

   dp slice is size n but code accesses dp[i] where i can be n, causing out of range error.

2. Step 2: Fix dp size to n+1

   Make dp size n+1 to safely store dp[n] and avoid index errors.

3. Final Answer:

   Change dp size to n+1 and access dp[i] safely -> Option A

4. Quick Check:

   dp size must be n+1 for dp[n] access [OK]

Hint: dp array size must be n+1 to avoid index errors [OK]

Common Mistakes:

• Using dp size n causes index out of range
• Changing loop start incorrectly
• Removing inner loop breaks logic

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. Max profit 12 by cuts [2,3]

B. Max profit 11 by cuts [1,4]

C. Max profit 13 by cuts [1,1,3]

D. Max profit 10 by cuts [5]

Solution

1. Step 1: Calculate max profit dp array

   dp[0] = 0
   dp[1] = max(prices[0] + dp[0]) = 2 + 0 = 2
   dp[2] = max(2 + dp[1] = 4, 5 + dp[0] = 5) = 5
   dp[3] = max(2 + dp[2] = 7, 5 + dp[1] = 7, 7 + dp[0] = 7) = 7
   dp[4] = max(2 + dp[3] = 9, 5 + dp[2] = 10, 7 + dp[1] = 9, 8 + dp[0] = 8) = 10
   dp[5] = max(2 + dp[4] = 12, 5 + dp[3] = 12, 7 + dp[2] = 12, 8 + dp[1] = 10, 10 + dp[0] = 10) = 12

2. Step 2: Identify cuts for max profit 12

   Max profit 12 achieved by cuts [2,3] or [3,2]. Prices 5 + 7 = 12.

3. Final Answer:

   Max profit 12 by cuts [2,3] -> Option A

4. Quick Check:

   Max profit 12 with cuts 2 and 3 [OK]

Hint: Calculate dp and track cuts for max profit [OK]

Common Mistakes:

• Not checking all cut combinations
• Assuming single cut is best
• Ignoring dp values for smaller lengths