What is the Rod Cutting Problem?
Learn the rod cutting dynamic programming problem, its O(n^2) bottom-up solution, and how to answer this interview question well.
Expected Interview Answer
The rod cutting problem asks for the maximum revenue obtainable by cutting a rod of length n into pieces and selling each piece at a given price, and it is solved with dynamic programming in O(n^2) time by building up the best revenue for every smaller rod length first.
For a rod of length n, you try every first cut length i from 1 to n, and the best revenue is price[i] plus the best revenue for the remaining rod of length n-i, which is itself a smaller instance of the same problem. Naive recursion recomputes the same sub-lengths exponentially many times, so a bottom-up table revenue[0..n] is built where revenue[j] is derived from all revenue[j-i] for i from 1 to j, each already computed. This gives O(n^2) time and O(n) space, and by tracking which cut length produced the best value at each step you can also reconstruct the actual cut sizes, not just the total. It is the canonical entry point into cut/partition style dynamic programming, directly generalizing to problems like coin change and matrix chain multiplication.
- O(n^2) time versus exponential brute force recursion
- Reconstructs the actual optimal cut sizes, not just the total value
- Reuses overlapping sub-length solutions instead of recomputing them
- Template for other cut and partition dynamic programming problems
AI Mentor Explanation
A groundskeeper has one long boundary rope and must cut it into segments to sell as souvenir lengths, where each specific segment length fetches a fixed price at the merchandise stand. Instead of guessing, they record the best possible sale value for every shorter rope length first, from one metre up to the full rope. For a rope of length n, they check every first-cut option i, add the price of that piece to the already-known best value for the remaining n-i metres, and keep the highest total. This bottom-up table means the value for a ten-metre rope is never recomputed twice, even though many different final cutting patterns share the same shorter sub-lengths.
Step-by-Step Explanation
Step 1
Define the subproblem
revenue[j] = max revenue obtainable from a rod of length j, for j from 0 to n.
Step 2
Base case
revenue[0] = 0, since a rod of length zero has no value.
Step 3
Fill bottom-up
For each j, try every first cut i (1..j) and set revenue[j] = max(price[i] + revenue[j-i]) over all i.
Step 4
Reconstruct cuts (optional)
Track which i achieved the max at each j to walk back and recover the actual optimal cut sizes.
What Interviewer Expects
- Identify the optimal substructure: best value for length n depends on best values for smaller lengths
- State the O(n^2) time and O(n) space bottom-up solution, not just naive recursion
- Explain why memoization or tabulation avoids exponential recomputation
- Mention that cut sizes can be reconstructed by tracking the choice at each length
Common Mistakes
- Writing only the naive exponential recursion without memoization
- Confusing rod cutting with the unbounded knapsack problem despite the structural similarity
- Forgetting the base case revenue[0] = 0
- Not realizing piece lengths can repeat (this is an unbounded, not 0/1, choice per length)
Best Answer (HR Friendly)
“The rod cutting problem is about finding the best way to slice up a rod into pieces to maximize the money you get from selling them, when each length has its own price. I solve it by building up the best answer for short rods first, then reusing those answers to solve longer rods, which avoids recalculating the same smaller cases over and over.”
Code Example
def rod_cutting(price, n):
# price[i] = value of a piece of length i (1-indexed conceptually)
revenue = [0] * (n + 1)
first_cut = [0] * (n + 1)
for length in range(1, n + 1):
best = float("-inf")
best_cut = length
for i in range(1, length + 1):
candidate = price[i] + revenue[length - i]
if candidate > best:
best = candidate
best_cut = i
revenue[length] = best
first_cut[length] = best_cut
# reconstruct the cuts
cuts = []
remaining = n
while remaining > 0:
cuts.append(first_cut[remaining])
remaining -= first_cut[remaining]
return revenue[n], cutsFollow-up Questions
- How does rod cutting relate to the unbounded knapsack problem?
- How would you reduce the time complexity using a different formulation?
- What changes if there is a fixed cost per cut?
- How would you solve this with top-down memoization instead of tabulation?
MCQ Practice
1. What is the time complexity of the standard bottom-up rod cutting solution for a rod of length n?
For each of the n lengths, up to n cut options are tried, giving O(n^2) total time.
2. What is the base case in the rod cutting dynamic programming table?
A rod of length zero yields zero revenue, which anchors the recurrence for all larger lengths.
3. Which problem is rod cutting most structurally similar to?
Rod cutting allows reusing the same piece length any number of times, matching the unbounded knapsack pattern.
Flash Cards
What does rod cutting optimize for? — Maximum total revenue from cutting a rod of length n given a price for each piece length.
What is the time complexity of the DP solution? — O(n^2), trying every first cut for every rod length.
How do you recover the actual cut sizes? — Track which first-cut choice produced the best value at each length, then walk back from n.
Is rod cutting more like 0/1 or unbounded knapsack? — Unbounded knapsack, since a piece length can be reused any number of times.