Rod Cutting Problem in DSA Go - Build from Scratch

Rod Cutting Problem

📖 Scenario: You have a metal 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 price you can get by cutting a rod into pieces using given prices for each length.

📋 What You'll Learn

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

Create a variable called rodLength with the value 8

Use a slice called maxPrices to store the best prices for each length

Use a for loop with variable length from 1 to rodLength

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

Print the maximum price for the rod length

💡 Why This Matters

🌍 Real World

This problem models real-life situations where you want to cut raw materials into pieces to maximize profit, like cutting metal rods, wood, or fabric.

💼 Career

Understanding dynamic programming and optimization problems is useful for software engineers working on resource allocation, manufacturing software, and financial modeling.

1

Create the prices slice

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

DSA Go

# Create the prices slice with given values
# Your code here

Hint

Use prices := []int{1, 5, 8, 9, 10, 17, 17, 20} to create the slice.

2

Set the rod length

Create a variable called rodLength and set it to 8

DSA Go

package main

func main() {
    prices := []int{1, 5, 8, 9, 10, 17, 17, 20}
    // Set rodLength to 8
    // Your code here
}

Hint

Use rodLength := 8 to set the rod length.

3

Calculate maximum prices for each length

Create a slice called maxPrices with length rodLength + 1 initialized with zeros. Use a for loop with variable length from 1 to rodLength. Inside it, use a nested for loop with variable cut from 1 to length. Update maxPrices[length] with the maximum of its current value and prices[cut-1] + maxPrices[length - cut].

DSA Go

package main

func main() {
    prices := []int{1, 5, 8, 9, 10, 17, 17, 20}
    rodLength := 8
    // Create maxPrices slice with zeros
    // Use for loops to calculate maxPrices
    // Your code here
}

Hint

Use make([]int, rodLength+1) to create the slice. Use nested loops to update maxPrices.

4

Print the maximum price

Use fmt.Println to print maxPrices[rodLength].

DSA Go

package main

import "fmt"

func main() {
    prices := []int{1, 5, 8, 9, 10, 17, 17, 20}
    rodLength := 8
    maxPrices := make([]int, rodLength+1)

    for length := 1; length  maxVal {
                maxVal = val
            }
        }
        maxPrices[length] = maxVal
    }
    // Print the maximum price for rodLength
    // Your code here
}

Hint

Use fmt.Println(maxPrices[rodLength]) to print the result.

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