Rod Cutting Problem in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Rod Cutting

What is the output of the following code that uses a simple recursive approach to solve the rod cutting problem?

DSA Python

def rod_cutting(prices, n):
    if n == 0:
        return 0
    max_val = float('-inf')
    for i in range(1, n+1):
        max_val = max(max_val, prices[i-1] + rod_cutting(prices, n - i))
    return max_val

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

A 9

B 10

C 8

D 5

💡 Hint

Think about all possible ways to cut the rod and sum prices.

The function tries all cuts and returns the maximum price. For length 4, the best is cutting into lengths 2 and 2 (5 + 5 = 10).

❓ Predict Output

intermediate

Output of Bottom-Up Rod Cutting

What is the output of this bottom-up dynamic programming solution for the rod cutting problem?

DSA Python

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

prices = [2, 5, 7, 8]
n = 4
print(rod_cutting_bottom_up(prices, n))

A 10

B 9

C 8

D 7

💡 Hint

Check how dp array builds up maximum prices for smaller lengths.

The best cut for length 4 is two pieces of length 2 (5 + 5 = 10).

🔧 Debug

advanced

Identify the Error in Memoized Rod Cutting

What error does the following memoized rod cutting code produce when run?

DSA Python

def rod_cutting_memo(prices, n, memo):
    if n == 0:
        return 0
    if n in memo:
        return memo[n]
    max_val = float('-inf')
    for i in range(1, n + 1):
        max_val = max(max_val, prices[i-1] + rod_cutting_memo(prices, n - i, memo))
    memo[n] = max_val
    return max_val

prices = [1, 5, 8, 9]
memo = {}
n = 4
print(rod_cutting_memo(prices, n, memo))

A 9

B 10

C IndexError

D RecursionError

💡 Hint

Check the loop range and how it affects the last piece.

The loop misses the case when i == n, so it does not consider the full length cut, resulting in a lower max value.

🧠 Conceptual

advanced

Time Complexity of Rod Cutting Algorithms

What is the time complexity of the naive recursive rod cutting solution without memoization for a rod of length n?

A O(n^2)

B O(n log n)

C O(n)

D O(2^n)

💡 Hint

Consider how many recursive calls are made for each length.

The naive recursive solution explores all possible cuts, leading to exponential calls.

🚀 Application

expert

Maximum Revenue and Cuts for Rod Cutting

Given the prices and rod length, which option correctly returns both the maximum revenue and the list of cuts to achieve it?

DSA Python

def rod_cutting_with_solution(prices, n):
    dp = [0] * (n + 1)
    cuts = [0] * (n + 1)
    for i in range(1, n + 1):
        max_val = float('-inf')
        for j in range(1, i + 1):
            if max_val < prices[j - 1] + dp[i - j]:
                max_val = prices[j - 1] + dp[i - j]
                cuts[i] = j
        dp[i] = max_val
    res = []
    length = n
    while length > 0:
        res.append(cuts[length])
        length -= cuts[length]
    return dp[n], res

prices = [1, 5, 8, 9]
n = 4
print(rod_cutting_with_solution(prices, n))

A (7, [1, 1, 1, 1])

B (9, [4])

C (10, [2, 2])

D (8, [3, 1])

💡 Hint

Trace the cuts array to find which pieces are chosen.

The maximum revenue is 10 by cutting the rod into two pieces of length 2 each.

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