Rod Cutting Problem in DSA Go

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 cuts 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, not just sell the whole bar as one piece.

Rod length: 8 units
Prices: [1:1, 2:5, 3:8, 4:9, 5:10, 6:17, 7:17, 8:20]

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

Cut options:
[1] + [7]
[2] + [6]
[3] + [5]
...
Try all to find max price

Dry Run Walkthrough

Input: rod length 4, prices: length 1=1, 2=5, 3=8, 4=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

Why: Selling whole rod is one option

Step 2: Try cut at length 1: price(1) + best price for length 3

price(1)=1 + best price(3) unknown yet

Why: We break rod into 1 and 3, need best price for 3

Step 3: Find best price for length 3: try no cut price(3)=8

best price(3) = 8 so far

Why: Selling length 3 whole is one option

Step 4: Try cut length 1 + length 2 for length 3: price(1)+price(2)=1+5=6

best price(3) remains 8 (max of 8 and 6)

Why: Check if cutting length 3 improves price

Step 5: Try cut length 2 + length 1 for length 3: price(2)+price(1)=5+1=6

best price(3) remains 8

Why: Check all cuts for length 3

Step 6: Calculate price for cut at length 1 for rod length 4: 1 + 8 = 9

Max price so far = max(9,9) = 9

Why: Compare with no cut price

Step 7: Try cut at length 2: price(2) + best price for length 2

price(2)=5 + best price(2) unknown

Why: Need best price for length 2

Step 8: Find best price for length 2: no cut price(2)=5

best price(2) = 5 so far

Why: Selling length 2 whole is one option

Step 9: Try cut length 1 + length 1 for length 2: price(1)+price(1)=1+1=2

best price(2) remains 5

Why: Check if cutting length 2 improves price

Step 10: Calculate price for cut at length 2 for rod length 4: 5 + 5 = 10

Max price so far = max(9,10) = 10

Why: Cutting at length 2 gives better price

Step 11: Try cut at length 3: price(3) + best price for length 1 = 8 + 1 = 9

Max price so far = max(10,9) = 10

Why: Check last cut option

Result:

Max price = 10 by cutting rod into two pieces of length 2 each
Final cuts: [2] + [2]

Annotated Code

DSA Go

package main

import "fmt"

// rodCutting returns max price for rod length n given prices array
func rodCutting(prices []int, n int) int {
	if n == 0 {
		return 0
	}
	maxVal := 0
	for i := 1; i <= n; i++ {
		// price for first cut i + best price for remaining n-i
		val := prices[i-1] + rodCutting(prices, n-i)
		if val > maxVal {
			maxVal = val
		}
	}
	return maxVal
}

func main() {
	prices := []int{1, 5, 8, 9}
	n := 4
	maxPrice := rodCutting(prices, n)
	fmt.Printf("Max price for rod length %d is %d\n", n, maxPrice)
}

if n == 0 { return 0 }

base case: no rod means no price

for i := 1; i <= n; i++ {

try all cut lengths from 1 to n

val := prices[i-1] + rodCutting(prices, n-i)

calculate price for cut i plus best price for remaining

if val > maxVal { maxVal = val }

update max price if current cut is better

return maxVal

return best price found

Output

Max price for rod length 4 is 10

Complexity Analysis

Time: O(2^n) because it tries all cuts recursively without remembering results

Space: O(n) due to recursion call stack depth up to n

vs Alternative: Dynamic programming approach reduces time to O(n^2) by storing intermediate results

Edge Cases

rod length 0

returns 0 price immediately

DSA Go

if n == 0 { return 0 }

rod length 1

returns price of length 1 piece

DSA Go

for i := 1; i <= n; i++ { ... }

prices array shorter than rod length

code assumes prices length >= rod length, otherwise index error

DSA Go

none (input must be valid)

When to Use This Pattern

When asked to maximize profit by cutting a rod or dividing a resource into parts with given prices, use the Rod Cutting pattern to try all partitions and pick the best.

Common Mistakes

Mistake: Not trying all possible cuts and only checking a few
Fix: Use a loop from 1 to n to try every cut length

Mistake: Forgetting base case n=0 causing infinite recursion
Fix: Add base case returning 0 when rod length is zero

Mistake: Not adding price of current cut piece before recursive call
Fix: Add prices[i-1] to recursive result for remaining length

Summary

It finds the best way to cut a rod to get maximum total price from selling pieces.

Use it when you want to maximize profit by cutting a resource into parts with known prices.

The key is to try all cut lengths and combine their prices recursively to find the best total.

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