Rod Cutting Problem | Easy Explanation & Code | Dynamic Programming

Rod Cutting Problem | Easy Explanation & Code | Dynamic Programming | DSA-One Course #92

Rod Cutting is a classic Dynamic Programming problem faced during technical interviews. It involves cutting a rod into multiple segments in such a way that the total revenue is maximized. The revenue from each piece is predefined and provided in a list as input.

Introduction

From time immemorial, individuals have faced the challenge of rod cutting to maximize profits. In this article, we'll delve deep into solving the rod cutting problem using dynamic programming techniques.

Problem Statement

Given a rod of length n and a set of prices for each possible length, the goal is to determine the maximum revenue obtainable by cutting up the rod and selling the pieces.

Example:

For instance, if the rod length is 4 and the prices for each length are:

Length 1: \$ 1
Length 2: \$ 5
Length 3: \$ 8
Length 4: \$ 9

The aim is to find the best way to cut the rod to maximize profit.

Approach

To solve this problem, we utilize a bottom-up dynamic programming approach. This involves storing the maximum revenue obtainable for each smaller length and then using those results to build up to the solution for the full length.

Here’s a breakdown of the steps:

Initialization: Start by creating an array to store the maximum revenue for each length.
Iterative Calculation: For each length, calculate the maximum revenue by considering all possible cuts and keeping track of the best option.
Final Solution: The value for the full length will give the required maximum revenue.

Dynamic Programming Solution:

Here’s the core logic to solve the problem in Python:

def rod_cutting(prices, n):
    # Create a revenue array to store the maximum revenue for each length
    dp = [0] * (n + 1)
    
    # Iterate over each length from 1 to n
    for i in range(1, n + 1):
        max_val = float('-inf')
        
        # Check all possible cuts and choose the best one
        for j in range(i):
            max_val = max(max_val, prices[j] + dp[i - j - 1])
        dp[i] = max_val
    
    return dp[n]

## Introduction
prices = [1, 5, 8, 9]
n = 4
print(f"Maximum revenue is: (rod_cutting(prices, n))")

Explanation:

Initialization: We start with a dp array of size n+1 filled with zeros.
Iterative Calculation: For each length i, we check all cuts j (from 0 to i-1) and update dp[i] with the maximum value obtained.
Final Solution: The solution for the rod of full length n is stored in dp[n].

Conclusion:

By dynamically calculating the maximum revenue for smaller lengths and building up, we efficiently solve the rod cutting problem. This approach ensures that we reuse previously computed results, optimizing the overall complexity.

Keywords

Rod Cutting
Dynamic Programming
Revenue Maximization
Technical Interviews
Python Code
Bottom-Up Approach

FAQs

Q: What is the rod cutting problem? A: The rod cutting problem involves cutting a rod of given length into segments to maximize total revenue, with predefined prices for each possible segment length.

Q: Why use Dynamic Programming for this problem? A: Dynamic Programming efficiently solves optimization problems by storing and reusing results of overlapping subproblems, which reduces computational redundancy.

Q: How does the bottom-up approach work in this context? A: The bottom-up approach involves solving and storing the solutions to smaller subproblems first, and using these solutions to address larger subproblems incrementally.

Q: Can this solution be applied to any revenue list and rod length? A: Yes, as long as the list of prices is provided and the length of the rod is specified, this approach can be used to determine the maximum obtainable revenue for those inputs.

Q: What is the time complexity of the provided solution? A: The time complexity of this dynamic programming solution is O(n^2), where n is the length of the rod.

Q: Is there a real-life application of the rod cutting problem? A: Yes, this problem can be applied in industries where materials must be cut and sold in segments, such as metalworking, woodworking, and textiles, to maximize profits.