Rod Cutting Problem in DSA Python

Mental Model

We want to cut a rod into pieces to get the highest total price by choosing the best lengths to cut.

Analogy: Imagine you have a chocolate bar and a price list for different sizes. You want to break it into pieces to sell for the most money.

Rod length: 8 units
Prices: [1, 5, 8, 9, 10, 17, 17, 20]
Rod: 8 units long -> [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
Each box can be cut and sold separately.

Dry Run Walkthrough

Input: rod length = 4, prices = [1, 5, 8, 9]

Goal: Find the best way to cut the rod of length 4 to get the maximum price.

Step 1: Calculate max price for rod length 1

Length 1: max price = 1

Why: Base case: only one piece, price is prices[0]

Step 2: Calculate max price for rod length 2 by checking cuts at 1 and 2

Length 2: max price = max(prices[1], prices[0]+prices[0]) = max(5, 1+1) = 5

Why: Either sell whole length 2 or two pieces of length 1

Step 3: Calculate max price for rod length 3 by checking cuts at 1, 2, and 3

Length 3: max price = max(prices[2], prices[0]+max_price(2), prices[1]+max_price(1)) = max(8, 1+5, 5+1) = 8

Why: Try all cuts to find best total price

Step 4: Calculate max price for rod length 4 by checking cuts at 1, 2, 3, and 4

Length 4: max price = max(prices[3], prices[0]+max_price(3), prices[1]+max_price(2), prices[2]+max_price(1)) = max(9, 1+8, 5+5, 8+1) = 10

Why: Try all cuts to maximize total price

Result:

Max price for rod length 4 is 10

Annotated Code

DSA Python

class RodCutting:
    def __init__(self, prices):
        self.prices = prices

    def max_profit(self, length):
        dp = [0] * (length + 1)  # dp[i] stores max price for rod length i
        for i in range(1, length + 1):
            max_val = float('-inf')
            for j in range(1, min(i, len(self.prices)) + 1):
                # Check if cutting at length j gives better price
                max_val = max(max_val, self.prices[j - 1] + dp[i - j])
            dp[i] = max_val
        return dp[length]

# Driver code
prices = [1, 5, 8, 9]
rod_length = 4
rc = RodCutting(prices)
print(rc.max_profit(rod_length))

for i in range(1, length + 1):

Iterate over all rod lengths from 1 to target length

for j in range(1, min(i, len(self.prices)) + 1):

Try all possible first cut lengths for current rod length i, limited by prices list length

max_val = max(max_val, self.prices[j - 1] + dp[i - j])

Update max price by comparing current best with price of first cut plus best for remaining length

dp[i] = max_val

Store best price found for rod length i

Output

10

Complexity Analysis

Time: O(n^2) because for each length up to n, we try all possible cuts up to that length

Space: O(n) because we store max prices for all lengths from 0 to n in dp array

vs Alternative: Naive recursive approach is exponential due to repeated calculations; dynamic programming reduces it to polynomial time by storing intermediate results

Edge Cases

rod length = 0

Returns 0 because no rod means no profit

DSA Python

dp = [0] * (length + 1)  # dp[0] = 0 handles zero length

rod length greater than prices list length

Code now safely handles rod lengths greater than prices list length by limiting cuts to available prices

DSA Python

for j in range(1, min(i, len(self.prices)) + 1): limits cuts to prices length

all prices are zero

Returns 0 as max profit since no piece has value

DSA Python

max_val initialized to -inf and updated with prices, so zero prices handled naturally

When to Use This Pattern

When asked to maximize profit by cutting a rod or dividing a resource into parts with given prices, use dynamic programming to try all cuts and store best results.

Common Mistakes

Mistake: Not storing intermediate results and recalculating max profit for same lengths repeatedly
Fix: Use a dp array to save max profits for all lengths to avoid repeated work

Mistake: Off-by-one errors when accessing prices array with cut length
Fix: Remember prices index is cut length minus one (prices[j-1])

Summary

Find the best way to cut a rod to get the highest total price from given piece prices.

Use when you want to maximize profit by cutting a rod or similar resource into pieces with known prices.

Try all cuts for each length and store the best result to avoid repeated calculations.

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