Rod Cutting Problem
A medium Dynamic Programming problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
- Platform
- GeeksforGeeks
The problem
Given a rod of length n and an array of prices for pieces of different lengths, determine the maximum revenue obtainable by cutting the rod into pieces and selling them.
Example 1
- Input
- n = 8, prices = [1,5,8,9,10,17,17,20]
- Output
- 22
- Why
- Cut the rod into pieces of length 2 and 6, giving revenue 5 + 17 = 22.
Example 2
- Input
- n = 4, prices = [1,5,8,9]
- Output
- 10
- Why
- Cut into two pieces of length 2 for revenue 5 + 5 = 10.
Example 3
- Input
- n = 1, prices = [10]
- Output
- 10
- Why
- The rod of length 1 is sold as-is for revenue 10.
Constraints
- 1 <= n <= 1000
- prices.length == n
- 1 <= prices[i] <= 10^6
How to think about it
Updated 2026-09-09Every cut divides the rod into a piece of length i (sold immediately for prices[i-1]) and a remaining rod of length len - i (which can be cut further). Because you can reuse cuts of the same length as many times as you like, this is unbounded knapsack where piece lengths are weights.
Approaches, worst first
Recursive cut branching
time O(2^n) · space O(n)
At remaining length len, try every possible initial cut of size i from 1 to len and recurse on len - i. Generates 2^(n-1) cutting patterns with massive overlapping subproblems.
Dynamic programming tabulationWrite this one
time O(n^2) · space O(n)
Create array dp of size n + 1 where dp[len] is maximum revenue for length len. For len from 1 to n, iterate cut from 1 to len: dp[len] = max(dp[len], prices[cut-1] + dp[len - cut]).
Where people lose marks · 3
- Off-by-one indexing between cut length (1-based) and price index (0-based prices[cut - 1]).
- Assuming a greedy price-per-unit-length heuristic works. A piece with slightly lower density might leave zero wasted length and produce a higher total than an indivisible high-density piece.
- Incorrectly sweeping cuts backward as if items could only be used once. Pieces can be reused indefinitely, so dp[len - cut] must reference the already updated values.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
Companies that have asked it
Tags taken from the problem's own GeeksforGeeks page — not a copied list.
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