Rod Cutting Problem in DSA C - Build from Scratch

Rod Cutting Problem

📖 Scenario: You have a rod of a certain length and a price list for different lengths of rods. You want to cut the rod into pieces to get the maximum total price.Imagine you have a wooden stick and a price chart for selling pieces of different sizes. Your goal is to decide how to cut the stick to earn the most money.

🎯 Goal: Build a program that calculates the maximum price you can get by cutting a rod into pieces using a simple approach.

📋 What You'll Learn

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

Create an integer variable called rod_length with the value 8.

Use a loop to calculate the maximum price for the rod length using the prices array.

Print the maximum price.

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding this problem helps in roles involving optimization, manufacturing, and resource management where maximizing value from limited resources is important.

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

Use int prices[] = { ... }; to create the array with the given values.

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

Use int rod_length = 8; to set the rod length.

3

Calculate the maximum price

Create an integer variable called max_price and set it to 0. Then use a for loop with variable i from 1 to rod_length inclusive. Inside the loop, update max_price to the maximum of its current value and prices[i - 1] + prices[rod_length - i].

DSA C

int prices[] = {1, 5, 8, 9, 10, 17, 17, 20};
int rod_length = 8;
// Calculate the maximum price
// Your code here

Hint

Use a loop from 1 to rod_length and update max_price if the sum of prices for two pieces is greater.

4

Print the maximum price

Use printf to print the text "Maximum price: " followed by the value of max_price and a newline.

DSA C

int prices[] = {1, 5, 8, 9, 10, 17, 17, 20};
int rod_length = 8;
int max_price = 0;
for (int i = 1; i  max_price) {
        max_price = current_price;
    }
}
// Print the maximum price
// Your code here

Hint

Use printf("Maximum price: %d\n", max_price); to print the result.

Practice

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

easy

A. To minimize the number of cuts made on the rod

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

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 highest 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 = maximize total price [OK]

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

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing the problem with counting combinations
• Assuming the rod length changes

2. Which of the following is the correct way to declare the price array for a rod of length 4 in C?

easy

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

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

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

D. int price[4] = (2, 5, 7, 8);

Solution

1. Step 1: Understand array indexing for rod prices

   Prices are usually stored with index representing length, so for length 1 to 4, index 0 is unused or zero.

2. Step 2: Check array declaration correctness

   int price[5] = {0, 2, 5, 7, 8}; declares array of size 5 with price[0]=0, then prices for lengths 1 to 4, which is correct.

3. Final Answer:

   int price[5] = {0, 2, 5, 7, 8}; -> Option C

4. Quick Check:

   Index 0 unused, prices start at 1 [OK]

Hint: Use size n+1 with zero at index 0 for convenience [OK]

Common Mistakes:

• Using parentheses instead of braces for array initialization
• Declaring array size smaller than needed
• Not including zero price at index 0

3. What will be the output of the following C code snippet for rod length 4?

int price[5] = {0, 2, 5, 7, 8};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 1; j <= i; j++) {
    int val = price[j] + dp[i - j];
    if (val > max_val) max_val = val;
  }
  dp[i] = max_val;
}
printf("%d", dp[n]);

medium

A. 10

B. 9

C. 8

D. 7

Solution

1. Step 1: Trace dp array computation for each length

   Calculate dp[1] to dp[4]: dp[1] = max(price[1]+dp[0]) = 2 dp[2] = max(price[1]+dp[1], price[2]+dp[0]) = max(2+2,5+0)=5 dp[3] = max(price[1]+dp[2], price[2]+dp[1], price[3]+dp[0]) = max(2+5,5+2,7+0)=7 dp[4] = max(price[1]+dp[3], price[2]+dp[2], price[3]+dp[1], price[4]+dp[0]) = max(2+7,5+5,7+2,8+0)=10

2. Step 2: Identify final dp[n] value

   dp[4] = 10 is printed.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Calculate dp bottom-up, check all cuts for max profit [OK]

Common Mistakes:

• Forgetting to consider all cut lengths j
• Using wrong indices for price or dp
• Not initializing dp array properly

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

int price[5] = {0, 2, 5, 7, 8};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 1; j < i; j++) {
    int val = price[j] + dp[i - j];
    if (val > max_val) max_val = val;
  }
  dp[i] = max_val;
}
printf("%d", dp[n]);

medium

A. max_val should be initialized to -1

B. dp array size is too small

C. price array initialization is incorrect

D. Inner loop should run till j <= i, not j < i

Solution

1. Step 1: Analyze inner loop range

   The inner loop runs from j=1 to j

2. Step 2: Correct the loop condition

   Changing inner loop to j <= i ensures all possible cuts including full length are considered.

3. Final Answer:

   Inner loop should run till j <= i, not j < i -> Option D

4. Quick Check:

   Include full length cut by using j <= i [OK]

Hint: Inner loop must include full length cut (j <= i) [OK]

Common Mistakes:

• Using j < i misses full rod cut
• Incorrect dp array size causes errors
• Initializing max_val to 0 is correct here

5. Given a rod of length 5 and prices price = {0, 2, 5, 7, 8, 10}, what is the maximum profit and the cuts to achieve it?
Use the rod cutting logic to find the best cuts and profit.

hard

A. Profit = 12, Cuts = [2, 3]

B. Profit = 11, Cuts = [1, 4]

C. Profit = 10, Cuts = [5]

D. Profit = 13, Cuts = [1, 1, 3]

Solution

1. Step 1: Calculate maximum profit using dp

   Calculate dp for lengths 1 to 5: 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 maximum profit

   Profit 12 achieved by cutting rod into lengths 2 and 3 (price[2]+price[3]=5+7=12).

3. Final Answer:

   Profit = 12, Cuts = [2, 3] -> Option A

4. Quick Check:

   Max profit 12 with cuts 2 and 3 [OK]

Hint: Check all cut pairs for max sum profit [OK]

Common Mistakes:

• Choosing no cuts or single cut only
• Not checking all combinations for max profit
• Confusing profit values with cut lengths