Rod Cutting Problem in DSA C++ - Execution Trace

Concept Flow - Rod Cutting Problem

Start with rod length n

↓

Try all cut lengths i from 1 to n

↓

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

↓

Keep max price found

↓

Store max price for length n

↓

Repeat for smaller lengths until 0

↓

Return max price for full rod length

We try all possible first cuts, combine their prices with best solutions for remaining rod, and pick the best overall price.

Execution Sample

DSA C++

int rodCutting(int n, vector<int>& prices) {
  vector<int> dp(n+1, 0);
  for (int len = 1; len <= n; len++) {
    int max_val = 0;
    for (int cut = 1; cut <= len; cut++) {
      max_val = max(max_val, prices[cut-1] + dp[len - cut]);
    }
    dp[len] = max_val;
  }
  return dp[n];
}

This code finds the maximum price for a rod of length n by trying all cuts and using dynamic programming.

Execution Table

Step	Operation	Current Length (len)	Cut Length (cut)	Price for Cut	dp Array State	Max Price for len
1	Initialize dp array	-	-	-	[0,0,0,0,0]	0
2	len=1, cut=1	1	1	prices[0]+dp[0]=1+0=1	[0,1,0,0,0]	1
3	len=2, cut=1	2	1	prices[0]+dp[1]=1+1=2	[0,1,2,0,0]	2
4	len=2, cut=2	2	2	prices[1]+dp[0]=5+0=5	[0,1,5,0,0]	5
5	len=3, cut=1	3	1	prices[0]+dp[2]=1+5=6	[0,1,5,6,0]	6
6	len=3, cut=2	3	2	prices[1]+dp[1]=5+1=6	[0,1,5,6,0]	6
7	len=3, cut=3	3	3	prices[2]+dp[0]=8+0=8	[0,1,5,8,0]	8
8	len=4, cut=1	4	1	prices[0]+dp[3]=1+8=9	[0,1,5,8,9]	9
9	len=4, cut=2	4	2	prices[1]+dp[2]=5+5=10	[0,1,5,8,10]	10
10	len=4, cut=3	4	3	prices[2]+dp[1]=8+1=9	[0,1,5,8,10]	10
11	len=4, cut=4	4	4	prices[3]+dp[0]=9+0=9	[0,1,5,8,10]	10
12	Return max price for full rod	4	-	-	[0,1,5,8,10]	10

💡 All lengths processed, dp[n] holds max price for rod length n

Variable Tracker

Variable	Start	After len=1	After len=2	After len=3	After len=4	Final
dp	[0,0,0,0,0]	[0,1,0,0,0]	[0,1,5,0,0]	[0,1,5,8,0]	[0,1,5,8,10]	[0,1,5,8,10]
len	0	1	2	3	4	4
max_val	0	1	5	8	10	10
cut	-	1	1-2	1-3	1-4	-

Key Moments - 3 Insights

Why do we add dp[len - cut] to prices[cut-1] when calculating max_val?

Because dp[len - cut] stores the best price for the remaining rod length after cutting 'cut' length. See execution_table rows 3-11 where max_val updates by adding prices for current cut plus dp for leftover.

Why does the dp array start with all zeros?

Because a rod of length 0 has price 0. This base case allows building solutions for larger lengths. See execution_table step 1 where dp is initialized.

Why do we try all cuts from 1 to len for each length len?

To find the best first cut that maximizes total price. Each cut splits the rod into two parts: cut length and remaining length. See execution_table steps 2-11 showing all cuts tried for each len.

Visual Quiz - 3 Questions

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

A 6

B 8

C 5

D 9

💡 Hint

Check the 'Max Price for len' column at step 7 in execution_table

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

A Step 11

B Step 8

C Step 9

D Step 10

💡 Hint

Look at the 'dp Array State' column in execution_table rows 8-11

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

A It would increase to 7

B It would stay 5

C It would decrease

D It would become 12

💡 Hint

Look at step 4 where max_val is prices[1] + dp[0], changing prices[1] affects max_val

Concept Snapshot

Rod Cutting Problem:
- Goal: Maximize price by cutting rod into pieces
- Use dp array where dp[len] = max price for length len
- For each length len, try all cuts i (1 to len)
- dp[len] = max(prices[i-1] + dp[len - i])
- Return dp[n] for full rod
- Bottom-up dynamic programming approach

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We use a dp array where dp[len] stores the best price for rod length len. For each length from 1 to n, we try all possible first cuts. For each cut, we add the price of that cut length to the best price of the remaining rod length from dp. We keep track of the maximum price found and store it in dp[len]. After filling dp for all lengths, dp[n] gives the maximum price for the full rod. This bottom-up approach avoids repeated calculations and efficiently finds the solution.

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