Rod Cutting Problem in DSA Python - 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

↓

Return max price for full rod

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

Execution Sample

DSA Python

prices = [0, 1, 5, 8, 9]
length = 4
max_price = 0
for i in range(1, length+1):
    max_price = max(max_price, prices[i] + prices[length - i])
print(max_price)

This code tries all cuts for rod length 4 and finds the maximum price by combining cut prices.

Execution Table

Step	Operation	Cut Length i	Price for cut i	Price for remaining length	Total Price	Max Price So Far
1	Try cut	1	1	8	9	9
2	Try cut	2	5	5	10	10
3	Try cut	3	8	1	9	10
4	Try cut	4	9	0	9	10
5	Done	-	-	-	-	10

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

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
max_price	0	9	10	10	10	10

Key Moments - 3 Insights

Why do we add prices[i] and prices[length - i]?

Because prices[i] is the price for the first cut of length i, and prices[length - i] is the best price for the leftover rod. Adding them gives total price for that cut choice (see execution_table rows 1-4).

Why do we update max_price only if total price is greater?

We want the maximum revenue possible, so we keep track of the highest total price found so far (see max_price column in execution_table).

Why do we try all cut lengths from 1 to n?

Because any cut length could give the best price. We must check all to find the maximum (see Step column in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the max_price after trying cut length 2?

A 13

B 10

C 8

D 9

💡 Hint

Check the Max Price So Far column at Step 2 in execution_table

At which step does the total price first reach 10?

A Step 2

B Step 1

C Step 3

D Step 4

💡 Hint

Look at the Total Price column in execution_table rows

If prices[3] was 10 instead of 8, what would be the max_price after Step 3?

A 13

B 18

C 11

D 10

💡 Hint

Recalculate total price at Step 3 using prices[3]=10 and prices[1]=1 from execution_table

Concept Snapshot

Rod Cutting Problem:
- Given rod length n and prices for lengths
- Try all cuts i from 1 to n
- Calculate price[i] + best price for n-i
- Keep max total price
- Result is max revenue for full rod

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We try every possible first cut length i, add its price to the best price for the leftover rod length n-i, and keep track of the maximum total price found. This process repeats until all cut lengths are checked. The final maximum price is the answer. The example code tries all cuts for rod length 4 with given prices and prints the maximum revenue 10. Key points include adding prices of the cut and leftover, updating max price only if better, and trying all cut lengths. The execution table shows step-by-step how max price updates with each cut tried.

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 the problem with counting ways
• Assuming shortest rod length is the goal

2. Which of the following is the correct way to initialize the DP array for rod cutting of length n in Python?

easy

A. dp = [0] * (n + 1)

B. dp = [0] * n

C. dp = [null] * n

D. dp = []

Solution

1. Step 1: Understand DP array size

   We need to store results for rod lengths from 0 to n, so array size must be n+1.

2. Step 2: Choose correct initialization

   Initializing with zeros for all lengths including 0 is done by dp = [0] * (n + 1).

3. Final Answer:

   dp = [0] * (n + 1) -> Option A

4. Quick Check:

   DP array size = n+1 with zeros [OK]

Hint: DP array size must cover 0 to n lengths [OK]

Common Mistakes:

• Using dp size n instead of n+1
• Initializing with null instead of 0
• Leaving dp empty without size

3. What will be the output of the following code snippet for rod length 4 and prices [1, 5, 8, 9]?

def rod_cutting(prices, n):
    dp = [0] * (n + 1)
    for i in range(1, n + 1):
        max_val = -1
        for j in range(i):
            max_val = max(max_val, prices[j] + dp[i - j - 1])
        dp[i] = max_val
    return dp[n]

print(rod_cutting([1, 5, 8, 9], 4))

medium

A. 9

B. 10

C. 8

D. 5

Solution

1. Step 1: Trace DP array calculation

   For n=4, prices=[1,5,8,9], calculate dp values: - dp[1] = max(1 + dp[0]) = 1 - dp[2] = max(1 + dp[1], 5 + dp[0]) = max(1+1,5+0)=5 - dp[3] = max(1 + dp[2], 5 + dp[1], 8 + dp[0]) = max(1+5,5+1,8+0)=8 - dp[4] = max(1 + dp[3], 5 + dp[2], 8 + dp[1], 9 + dp[0]) = max(1+8,5+5,8+1,9+0)=max(9,10,9,9)=10

2. Step 2: Return final dp value

   The function returns dp[4] = 10, which is the maximum profit.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Check all cuts and pick max sum for each length [OK]

Common Mistakes:

• Ignoring smaller cuts and taking only full length price
• Off-by-one errors in indexing prices or dp
• Returning dp[n-1] instead of dp[n]

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

def rod_cut(prices, n):
    dp = [0] * n
    for i in range(1, n + 1):
        max_val = 0
        for j in range(i):
            max_val = max(max_val, prices[j] + dp[i - j - 1])
        dp[i] = max_val
    return dp[n]

print(rod_cut([2, 5, 7, 8], 4))

medium

A. max_val should start from -1

B. Loop range should start from 0, not 1

C. dp array size is incorrect, should be n+1

D. prices indexing is off by one

Solution

1. Step 1: Check dp array initialization

   dp is initialized with size n, but dp[i] accesses index n, causing IndexError.

2. Step 2: Correct dp size

   dp should be size n+1 to store results for lengths 0 to n.

3. Final Answer:

   dp array size is incorrect, should be n+1 -> Option C

4. Quick Check:

   dp size = n+1 to avoid index errors [OK]

Hint: DP array must cover all lengths including zero [OK]

Common Mistakes:

• Using dp size n instead of n+1
• Starting max_val from 0 instead of -1 (not critical here)
• Incorrect loop ranges causing index errors

5. Given a rod of length 5 and prices [2, 5, 7, 8, 10], what is the maximum profit and the cuts to achieve it? Choose the correct maximum profit and one valid cut combination.

hard

A. Maximum profit: 10; Cuts: [5]

B. Maximum profit: 14; Cuts: [1, 4]

C. Maximum profit: 13; Cuts: [3, 2]

D. Maximum profit: 12; Cuts: [2, 3]

Solution

1. Step 1: Calculate max profit using DP

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

2. Step 2: Identify cuts for max profit 12

   One valid cut is [2, 3] because price[2]=5 and price[3]=7, total 12.

3. Step 3: Check if better cuts exist

   Check cut [1,4]: price[1]=2 + price[4]=8 = 10, less than 12. Check cut [3,2]: price[3]=7 + price[2]=5 = 12. Check cut [5]: price[5]=10. No combination yields 14.

4. Final Answer:

   Maximum profit: 12; Cuts: [2, 3] -> Option D

5. Quick Check:

   Max profit 12 from cuts 2 and 3 [OK]

Hint: Try all cut pairs to find max sum [OK]

Common Mistakes:

• Assuming full length price is always max
• Ignoring multiple cut combinations
• Mixing up prices and lengths