Rod Cutting Problem in DSA C - Execution Trace

Concept Flow - Rod Cutting Problem

Start with full rod length N

↓

Try all cut lengths i from 1 to N

↓

Calculate price for cut length i + best price for remaining length N-i

↓

Store max price for length N

↓

Repeat for all lengths from 1 to N

↓

Result: max price for full rod length

We try all possible first cuts, combine their prices with the best prices for the leftover rod, and store the maximum price for each rod length.

Execution Sample

DSA C

int rodCut(int length, int prices[]) {
  int dp[length+1];
  dp[0] = 0;
  for (int i=1; i<=length; i++) {
    int max_val = 0;
    for (int j=1; j<=i; j++) {
      max_val = (max_val > prices[j-1] + dp[i-j] ? max_val : prices[j-1] + dp[i-j]);
    }
    dp[i] = max_val;
  }
  return dp[length];
}

This code finds the maximum price for a rod of given length by trying all cuts and using dynamic programming to store best prices.

Execution Table

Step	Operation	Current Length i	Cut Length j	Calculation	Max Price for i	dp Array State	Visual State
1	Initialize dp[0]	0	-	-	0	[0]	[]
2	Calculate max price for length 1	1	1	prices[0] + dp[0] = 1 + 0 = 1	1	[0,1]	[1]
3	Calculate max price for length 2	2	1	prices[0] + dp[1] = 1 + 1 = 2	2	[0,1,2]	[2]
4	Calculate max price for length 2	2	2	prices[1] + dp[0] = 5 + 0 = 5	5	[0,1,5]	[5]
5	Calculate max price for length 3	3	1	prices[0] + dp[2] = 1 + 5 = 6	6	[0,1,5,6]	[6]
6	Calculate max price for length 3	3	2	prices[1] + dp[1] = 5 + 1 = 6	6	[0,1,5,6]	[6]
7	Calculate max price for length 3	3	3	prices[2] + dp[0] = 8 + 0 = 8	8	[0,1,5,8]	[8]
8	Calculate max price for length 4	4	1	prices[0] + dp[3] = 1 + 8 = 9	9	[0,1,5,8,9]	[9]
9	Calculate max price for length 4	4	2	prices[1] + dp[2] = 5 + 5 = 10	10	[0,1,5,8,10]	[10]
10	Calculate max price for length 4	4	3	prices[2] + dp[1] = 8 + 1 = 9	10	[0,1,5,8,10]	[10]
11	Calculate max price for length 4	4	4	prices[3] + dp[0] = 9 + 0 = 9	10	[0,1,5,8,10]	[10]
12	Calculate max price for length 5	5	1	prices[0] + dp[4] = 1 + 10 = 11	11	[0,1,5,8,10,11]	[11]
13	Calculate max price for length 5	5	2	prices[1] + dp[3] = 5 + 8 = 13	13	[0,1,5,8,10,13]	[13]
14	Calculate max price for length 5	5	3	prices[2] + dp[2] = 8 + 5 = 13	13	[0,1,5,8,10,13]	[13]
15	Calculate max price for length 5	5	4	prices[3] + dp[1] = 9 + 1 = 10	13	[0,1,5,8,10,13]	[13]
16	Calculate max price for length 5	5	5	prices[4] + dp[0] = 10 + 0 = 10	13	[0,1,5,8,10,13]	[13]
17	Return max price for full length	5	-	-	13	[0,1,5,8,10,13]	[13]

💡 All lengths from 1 to 5 processed, dp array fully computed, maximum price for rod length 5 is 13

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 7	After Step 11	After Step 16	Final
dp	[0]	[0,1]	[0,1,5]	[0,1,5,8]	[0,1,5,8,10]	[0,1,5,8,10,13]	[0,1,5,8,10,13]
max_val	0	1	5	8	10	13	13
i (current length)	0	1	2	3	4	5	5
j (cut length)	-	1	2	3	4	5	-

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, not just one piece. The execution_table rows 3-16 show how each cut length j is tried to find the maximum price.

Why do we add prices[j-1] + dp[i-j] in the calculation?

prices[j-1] is the price for the first cut of length j, and dp[i-j] is the best price for the remaining rod length i-j. Adding them gives the total price for that cut combination, as shown in the Calculation column in execution_table.

Why does dp[0] start at 0?

dp[0] represents the price for a rod of length 0, which is zero because no rod means no price. This base case allows building up solutions for larger lengths, as seen in Step 1 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what is the max price for rod length 3?

A 8

B 5

C 6

D 9

💡 Hint

Check the Max Price for i column at Step 7 in execution_table

At which step does the dp array first contain the value 10?

A Step 11

B Step 9

C Step 10

D Step 8

💡 Hint

Look at the dp Array State column in execution_table rows around Step 9 and Step 11

If prices[1] changed from 5 to 7, how would max price for rod length 2 change at Step 4?

A It would remain 5

B It would become 2

C It would become 7

D It would become 12

💡 Hint

Look at Calculation in Step 4: prices[1] + dp[0], changing prices[1] affects max price for length 2

Concept Snapshot

Rod Cutting Problem:
- Goal: Maximize price by cutting rod into pieces
- Use dp array where dp[i] = max price for length i
- For each length i, try all cuts j (1 to i)
- dp[i] = max(prices[j-1] + dp[i-j])
- dp[0] = 0 base case
- Result: dp[length] is max price

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to get the highest price. We use a dp array where each position stores the best price for that rod length. Starting from length 0 with price 0, we build up to the full rod length. For each length i, we try all possible first cuts j and calculate the total price as prices[j-1] plus the best price for the leftover rod dp[i-j]. We keep the maximum of these values in dp[i]. After filling dp for all lengths, dp[length] gives the maximum price. The execution table shows each step, the calculations, and how dp array changes. Key points include trying all cuts, adding prices and dp values, and starting dp[0] at zero. The visual quiz tests understanding of dp values at specific steps and effects of price changes.

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