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

↓

Keep max price found

↓

Store max price for length N

↓

Repeat for smaller lengths until length 0

↓

Return max price for full rod length

We try all possible first cuts, combine their prices with best prices for the leftover rod, and pick the best total price.

Execution Sample

DSA C#

int RodCut(int[] prices, int n) {
  if (n == 0) return 0;
  int max_val = int.MinValue;
  for (int i = 1; i <= n; i++) {
    max_val = Math.Max(max_val, prices[i-1] + RodCut(prices, n - i));
  }
  return max_val;
}

This code finds the best price for a rod of length n by trying all first cuts and recursively solving the rest.

Execution Table

Step	Operation	Rod Length (n)	Cut Length (i)	Price for cut	Remaining length	Best price so far	Visual State (Max prices array)
1	Start	4	-	-	-	-∞	[0, -, -, -, -]
2	Try cut i=1	4	1	2	3	?	[0, -, -, -, -]
3	Recurse for length 3	3	-	-	-	-	[0, -, -, -, -]
4	Try cut i=1	3	1	2	2	?	[0, -, -, -, -]
5	Recurse for length 2	2	-	-	-	-	[0, -, -, -, -]
6	Try cut i=1	2	1	2	1	?	[0, -, -, -, -]
7	Recurse for length 1	1	-	-	-	-	[0, -, -, -, -]
8	Try cut i=1	1	1	2	0	2	[0, 2, -, -, -]
9	Return max price for length 1	1	-	-	-	2	[0, 2, -, -, -]
10	Try cut i=2	2	2	5	0	5	[0, 2, 5, -, -]
11	Return max price for length 2	2	-	-	-	5	[0, 2, 5, -, -]
12	Try cut i=2	3	2	5	1	7	[0, 2, 5, 7, -]
13	Try cut i=3	3	3	7	0	7	[0, 2, 5, 7, -]
14	Return max price for length 3	3	-	-	-	7	[0, 2, 5, 7, -]
15	Try cut i=3	4	3	7	1	10	[0, 2, 5, 7, 10]
16	Try cut i=4	4	4	8	0	10	[0, 2, 5, 7, 10]
17	Return max price for length 4	4	-	-	-	10	[0, 2, 5, 7, 10]

💡 All cuts for rod length 4 tried; max price found is 10

Variable Tracker

Variable	Start	After Step 8	After Step 11	After Step 14	After Step 17
n	4	1	2	3	4
i	-	1	2	3	4
max_val	-∞	2	5	7	10
max_prices	[0, -, -, -, -]	[0, 2, -, -, -]	[0, 2, 5, -, -]	[0, 2, 5, 7, -]	[0, 2, 5, 7, 10]

Key Moments - 3 Insights

Why do we try all cut lengths from 1 to n instead of just one?

Because the best price might come from cutting the rod into smaller pieces, not just one big piece. See execution_table steps 2-16 where each cut length is tried and max price updated.

Why do we add prices[i-1] to the recursive call result?

prices[i-1] is the price for the first cut piece of length i, and the recursive call finds the best price for the leftover rod of length n-i. Adding them gives total price for that cut choice (see execution_table columns 'Price for cut' and 'Remaining length').

How do we know when to stop recursion?

When rod length n is 0, no more cuts can be made, so price is 0 (see code line 'if (n == 0) return 0;' and execution_table step 8 where remaining length is 0).

Visual Quiz - 3 Questions

Look at the execution_table at step 11, what is the max price found for rod length 2?

A 2

B 7

C 5

D 9

💡 Hint

Check the 'Best price so far' column at step 11 in execution_table.

At which step does the recursion return the max price for rod length 3?

A Step 11

B Step 14

C Step 9

D Step 17

💡 Hint

Look for 'Return max price for length 3' in the Operation column of execution_table.

If the price for length 4 was increased to 10, how would the max price for rod length 4 change at step 17?

A It would become 10

B It would become 11

C It would remain 9

D It would become 8

💡 Hint

Check step 16 where cut length 4 price is 8; increasing it to 10 would update max price at step 17.

Concept Snapshot

Rod Cutting Problem:
- Given rod length n and prices for each length,
- Try all cuts i=1 to n,
- Calculate price[i-1] + best price for remaining n-i,
- Keep max price found,
- Return max price for full rod length.
- Uses recursion or DP for efficiency.

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We try all possible first cuts, add their prices to the best prices for the leftover rod, and pick the maximum. The recursion stops when the rod length is zero. The execution table shows step-by-step how the max prices are computed for rod lengths 1 to 4. Variable tracking shows how max prices update after each step. Key moments clarify why all cuts are tried, why prices add up, and when recursion stops. The quiz tests understanding of max prices at different steps and effects of price changes.

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 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 = To maximize the total profit by cutting the rod into pieces [OK]

Hint: Focus on maximizing profit from pieces, not cuts or counts [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 C# method signature for a function solving the rod cutting problem recursively?

easy

A. int CutRod(int prices, int[] n)

B. void CutRod(int prices[], int n)

C. int CutRod(int[] prices, int n)

D. int CutRod(int[] prices, int[] lengths)

Solution

1. Step 1: Check parameter types and order

   The function should take an array of prices and an integer length n.

2. Step 2: Verify return type

   The function returns an integer representing maximum profit.

3. Final Answer:

   int CutRod(int[] prices, int n) -> Option C

4. Quick Check:

   Correct signature returns int and takes int[] and int [OK]

Hint: Return int max profit; input array prices and int length [OK]

Common Mistakes:

• Using void instead of int return type
• Swapping parameter types
• Passing arrays incorrectly

3. What is the output of the following C# code snippet for rod cutting?

int[] prices = {1, 5, 8, 9};
int n = 4;
int result = CutRod(prices, n);
Console.WriteLine(result);

int CutRod(int[] price, int length) {
    if (length == 0) return 0;
    int max_val = int.MinValue;
    for (int i = 1; i <= length; i++) {
        max_val = Math.Max(max_val, price[i - 1] + CutRod(price, length - i));
    }
    return max_val;
}

medium

A. 10

B. 9

C. 8

D. 5

Solution

1. Step 1: Understand recursive calls and prices

   Prices: length 1=1, 2=5, 3=8, 4=9. The function tries all cuts and sums prices recursively.

2. Step 2: Calculate max profit for length 4

   Possible cuts: - 1 + CutRod(3) = 1 + 8 = 9 - 5 + CutRod(2) = 5 + 5 = 10 - 8 + CutRod(1) = 8 + 1 = 9 - 9 + CutRod(0) = 9 + 0 = 9 Max is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Try all cuts, pick max sum of price + recursive result [OK]

Common Mistakes:

• Ignoring recursive calls
• Choosing first price only
• Misindexing price array

4. Identify the bug in this C# code for rod cutting and select the fix:

int CutRod(int[] price, int length) {
    if (length == 0) return 0;
    int max_val = 0;
    for (int i = 1; i <= length; i++) {
        max_val = Math.Max(max_val, price[i] + CutRod(price, length - i));
    }
    return max_val;
}

medium

A. Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds

B. Initialize max_val to int.MinValue instead of 0

C. Change loop to 'for (int i = 0; i < length; i++)' to fix indexing

D. Add base case for length == 1 to return price[0]

Solution

1. Step 1: Check array indexing in loop

   Loop uses i from 1 to length, but price array is zero-indexed. Accessing price[i] causes out of bounds when i == length.

2. Step 2: Fix indexing by subtracting 1

   Change price[i] to price[i - 1] to access correct element safely.

3. Final Answer:

   Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds -> Option A

4. Quick Check:

   Index fix = Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds [OK]

Hint: Array index starts at 0, loop starts at 1, adjust index [OK]

Common Mistakes:

• Using price[i] directly causing runtime error
• Changing loop incorrectly breaking logic
• Ignoring base case

5. Given a rod of length 5 and prices int[] prices = {2, 5, 7, 8, 10};, what is the maximum profit using dynamic programming bottom-up approach?

hard

A. 10

B. 12

C. 15

D. 14

Solution

1. Step 1: Initialize dp array for lengths 0 to 5

   dp[0]=0. Calculate dp[i] for i=1 to 5 using max of price[j-1]+dp[i-j].

2. Step 2: Calculate dp values step-by-step

   dp[1]=2 dp[2]=max(5,2+2)=5 dp[3]=max(7,5+2,2+5)=7 dp[4]=max(8,7+2,5+5,2+7)=10 dp[5]=max(10,8+2,7+5,5+7,2+10)=14

3. Final Answer:

   14 -> Option D

4. Quick Check:

   Bottom-up dp max profit = 14 [OK]

Hint: Build dp array from length 1 to n, pick max at each step [OK]

Common Mistakes:

• Not using dp array, causing repeated calculations
• Incorrect max calculation in loop
• Ignoring smaller cuts combinations