Rod Cutting Problem in DSA C++ - 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 length has a price. Your goal is to find the best way to cut the rod to get the most money.

🎯 Goal: Build a program that calculates the maximum money you can get by cutting the rod into pieces using given prices for each length.

📋 What You'll Learn

Create an array called prices with prices for rod lengths 1 to 8.

Create an integer variable rod_length set to 8.

Use a loop to find the maximum revenue for the rod of length rod_length.

Print the maximum revenue.

💡 Why This Matters

🌍 Real World

This problem models how to maximize profit when selling materials cut into pieces, like wood, metal rods, or fabric.

💼 Career

Understanding this problem helps in optimization tasks and dynamic programming, which are common in software engineering and algorithm design roles.

1

Create the prices array

Create an integer array called prices with these exact values: {1, 5, 8, 9, 10, 17, 17, 20}.

DSA C++

# Create the prices array
// Your code here

Hint

The array prices holds the price for each rod length from 1 to 8.

2

Set the rod length

Create an integer variable called rod_length and set it to 8.

DSA C++

int prices[] = {1, 5, 8, 9, 10, 17, 17, 20};
// Set the rod length
// Your code here

Hint

The variable rod_length is the total length of the rod you want to cut.

3

Calculate maximum revenue

Create an integer array called max_revenue of size rod_length + 1 and initialize all elements to 0. Then use a for loop with variable i from 1 to rod_length to find the maximum revenue for each length. Inside it, use another for loop with variable j from 1 to i to update max_revenue[i] with the maximum of its current value and prices[j-1] + max_revenue[i-j].

DSA C++

int prices[] = {1, 5, 8, 9, 10, 17, 17, 20};
int rod_length = 8;
// Calculate max revenue using loops
// Your code here

Hint

Use two loops: outer loop for rod lengths, inner loop for cutting positions. Update max revenue for each length.

4

Print the maximum revenue

Use std::cout to print the maximum revenue for the rod of length rod_length stored in max_revenue[rod_length].

DSA C++

int prices[] = {1, 5, 8, 9, 10, 17, 17, 20};
int rod_length = 8;
int max_revenue[rod_length + 1] = {0};
for (int i = 1; i  max_val) {
            max_val = val;
        }
    }
    max_revenue[i] = max_val;
}
// Print the maximum revenue
// Your code here

Hint

Print the value stored in max_revenue[rod_length] using std::cout.

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:

   Maximize profit = maximize total price [OK]

Hint: Focus on maximizing profit, not minimizing cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing with counting ways to cut
• Assuming shortest rod length is the goal

2. Which of the following is the correct way to declare the price array for a rod of length 5 in C++?

easy

A. int price[4] = {2, 5, 7, 8, 10};

B. int price[] = {2, 5, 7, 8, 10};

C. int price[6] = {0, 2, 5, 7, 8, 10};

D. int price[5] = (2, 5, 7, 8, 10);

Solution

1. Step 1: Understand array declaration in C++

   When size is omitted, the compiler infers it from initializer list length.

2. Step 2: Check each option

   int price[4] = {2, 5, 7, 8, 10}; declares size 4 but initializes 5 elements - invalid syntax (too many initializers). int price[] = {2, 5, 7, 8, 10}; omits size, compiler infers size 5 - correct and flexible. int price[6] = {0, 2, 5, 7, 8, 10}; declares size 6 but initializes 6 elements including 0 at start - usually price array starts at index 0 or 1, but this is uncommon. int price[5] = (2, 5, 7, 8, 10); uses parentheses instead of braces - invalid syntax.

3. Final Answer:

   int price[] = {2, 5, 7, 8, 10}; -> Option B

4. Quick Check:

   Array size inferred from initializer [OK]

Hint: Use braces {} for array initialization, omit size for flexibility [OK]

Common Mistakes:

• Using parentheses instead of braces
• Mismatching array size and initializer count
• Including extra zero element unnecessarily

3. What is the output of the following C++ code snippet for rod length 4?

int price[] = {1, 5, 8, 9};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 0; j < i; j++) {
    max_val = std::max(max_val, price[j] + dp[i - j - 1]);
  }
  dp[i] = max_val;
}
std::cout << dp[n] << std::endl;

medium

A. 10

B. 9

C. 8

D. 11

Solution

1. Step 1: Trace dp array calculation for n=4

   Calculate dp[1] to dp[4]:
   dp[1] = max(price[0]+dp[0]) = 1+0=1
   dp[2] = max(price[0]+dp[1], price[1]+dp[0]) = max(1+1, 5+0)=5
   dp[3] = max(price[0]+dp[2], price[1]+dp[1], price[2]+dp[0]) = max(1+5, 5+1, 8+0)= max(6,6,8)=8
   dp[4] = max(price[0]+dp[3], price[1]+dp[2], price[2]+dp[1], price[3]+dp[0]) = max(1+8, 5+5, 8+1, 9+0) = max(9,10,9,9)=10

2. Step 2: Verify final dp value

   dp[4] = 10, so output is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   dp[4] = 10 from max cuts [OK]

Hint: Calculate dp values stepwise from smaller lengths [OK]

Common Mistakes:

• Off-by-one errors in indexing
• Ignoring dp array values in calculation
• Confusing price array indexing

4. Identify the error in this C++ code snippet for rod cutting:

int price[] = {2, 5, 7, 8};
int n = 4;
int dp[4] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 0; j < i; j++) {
    max_val = std::max(max_val, price[j] + dp[i - j - 1]);
  }
  dp[i] = max_val;
}
std::cout << dp[n] << std::endl;

medium

A. std::max function is used incorrectly

B. Price array is missing an element for length 4

C. Loop variables i and j are swapped incorrectly

D. Array dp size is too small, causing out-of-bounds access

Solution

1. Step 1: Check dp array size and indexing

   dp is declared with size 4 but accessed at dp[i] where i goes up to 4, causing out-of-bounds access since valid indices are 0 to 3.

2. Step 2: Confirm other parts are correct

   Price array has 4 elements matching rod length. Loop variables and std::max usage are correct.

3. Final Answer:

   Array dp size is too small, causing out-of-bounds access -> Option D

4. Quick Check:

   dp size must be n+1 to avoid overflow [OK]

Hint: Declare dp array size as n+1 to avoid index errors [OK]

Common Mistakes:

• Using dp array size equal to n instead of n+1
• Ignoring array bounds in loops
• Misusing std::max parameters

5. Given a rod of length 5 and prices price = {2, 5, 7, 8, 10}, what is the maximum profit and the cuts to achieve it?

hard

A. Maximum profit 14 by cuts: 2 + 2 + 1

B. Maximum profit 13 by cuts: 1 + 4

C. Maximum profit 12 by cuts: 2 + 3

D. Maximum profit 15 by cuts: 1 + 1 + 3

Solution

1. Step 1: Calculate maximum profit using dp

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

2. Step 2: Check cuts for maximum profit

   Maximum profit 14 by cuts: 2 + 2 + 1 says profit 14 with cuts 2+2+1. Let's verify:
   Cut 2: price 5
   Cut 2: price 5
   Cut 1: price 2
   Total = 5+5+2=12, not 14.
   Check Maximum profit 15 by cuts: 1 + 1 + 3: 1+1+3 cuts:
   1:2,1:2,3:7 total=11
   Maximum profit 13 by cuts: 1 + 4: 1+4 cuts:
   1:2,4:8 total=10
   Maximum profit 12 by cuts: 2 + 3: 2+3 cuts:
   2:5,3:7 total=12
   So maximum profit is 12, cuts can be 2+3 or 2+2+1.
   Maximum profit 14 by cuts: 2 + 2 + 1 says 14 profit which is incorrect.
   Maximum profit 12 by cuts: 2 + 3 says 12 profit with cuts 2+3 which is correct.

3. Final Answer:

   Maximum profit 12 by cuts: 2 + 3 -> Option C

4. Quick Check:

   Max profit = 12 by cuts 2+3 [OK]

Hint: Try all cut combinations to find max profit [OK]

Common Mistakes:

• Adding prices incorrectly for cuts
• Assuming single cut gives max profit
• Ignoring multiple cut combinations