Rod Cutting Problem in DSA Javascript - Build from Scratch

Rod Cutting Problem

📖 Scenario: You have a metal rod of a certain length. You want to cut it into smaller pieces to sell. Each piece has a price depending on its length. Your goal is to find the best way to cut the rod to get the most money.

🎯 Goal: Build a program that calculates the maximum money you can get by cutting the rod into pieces using given prices for each length.

📋 What You'll Learn

Create an array called prices with prices for rod lengths 1 to 8.

Create a variable called rodLength with the value 8.

Use a loop to calculate the maximum revenue for each rod length from 1 to rodLength.

Print the maximum revenue for the rod of length 8.

💡 Why This Matters

🌍 Real World

This problem models how to cut raw materials like metal rods or wood to maximize profit, common in manufacturing and sales.

💼 Career

Understanding dynamic programming and optimization helps in software roles involving algorithms, operations research, and resource management.

1

Create the prices array

Create an array called prices with these exact values: [1, 5, 8, 9, 10, 17, 17, 20].

DSA Javascript

// Create the prices array
// Your code here

Hint

Use const prices = [1, 5, 8, 9, 10, 17, 17, 20]; to create the array.

2

Set the rod length

Create a variable called rodLength and set it to 8.

DSA Javascript

const prices = [1, 5, 8, 9, 10, 17, 17, 20];
// Create the rodLength variable
// Your code here

Hint

Use const rodLength = 8; to set the rod length.

3

Calculate maximum revenue

Create an array called maxRevenue with length rodLength + 1 filled with zeros. Then use a for loop with variable i from 1 to rodLength. Inside it, use another for loop with variable j from 1 to i. Update maxRevenue[i] with the maximum of its current value and prices[j - 1] + maxRevenue[i - j].

DSA Javascript

const prices = [1, 5, 8, 9, 10, 17, 17, 20];
const rodLength = 8;
// Calculate maxRevenue array and fill it
// Your code here

Hint

Use two nested loops and update maxRevenue[i] with the best price found.

4

Print the maximum revenue

Print the value of maxRevenue[rodLength] using console.log.

DSA Javascript

const prices = [1, 5, 8, 9, 10, 17, 17, 20];
const rodLength = 8;
const maxRevenue = new Array(rodLength + 1).fill(0);
for (let i = 1; i <= rodLength; i++) {
    for (let j = 1; j <= i; j++) {
        maxRevenue[i] = Math.max(maxRevenue[i], prices[j - 1] + maxRevenue[i - j]);
    }
}
// Print the maximum revenue
// Your code here

Hint

Use console.log(maxRevenue[rodLength]); to print the result.

Practice

1. What is the main goal of the Rod Cutting Problem in algorithms?

easy

A. To count the total number of ways to cut the rod

B. To minimize the number of cuts made on the rod

C. To find the shortest rod length possible

D. To maximize the total profit by cutting the rod into pieces

Solution

1. Step 1: Understand the problem objective

   The rod cutting problem aims to cut a rod into pieces to get the highest 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 D

4. Quick Check:

   Maximize profit = To maximize the total profit by cutting the rod into pieces [OK]

Hint: Remember: maximize profit, not minimize cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing number of ways with maximum profit
• Assuming rod length is to be minimized

2. Which of the following is the correct way to declare a function in JavaScript to solve the rod cutting problem?

easy

A. function rodCutting(prices, n) { /* code */ }

B. def rodCutting(prices, n): /* code */

C. func rodCutting(prices, n) { /* code */ }

D. rodCutting function(prices, n) { /* code */ }

Solution

1. Step 1: Identify JavaScript function syntax

   JavaScript functions are declared using the function keyword followed by the name and parameters.

2. Step 2: Check each option

   function rodCutting(prices, n) { /* code */ } uses correct JavaScript syntax. def rodCutting(prices, n): /* code */ is Python, C is Go/Swift style, D is invalid.

3. Final Answer:

   function rodCutting(prices, n) { /* code */ } -> Option A

4. Quick Check:

   JS function syntax = function rodCutting(prices, n) { /* code */ } [OK]

Hint: JS functions start with 'function' keyword [OK]

Common Mistakes:

• Using Python or other language syntax
• Missing 'function' keyword
• Incorrect parameter declaration

3. What will be the output of the following code snippet?

const prices = [1, 5, 8, 9];
function rodCutting(prices, n) {
  if (n === 0) return 0;
  let maxVal = -Infinity;
  for (let i = 1; i <= n; i++) {
    maxVal = Math.max(maxVal, prices[i - 1] + rodCutting(prices, n - i));
  }
  return maxVal;
}
console.log(rodCutting(prices, 4));

medium

A. 8

B. 9

C. 10

D. 15

Solution

1. Step 1: Understand the recursive calls for n=4

   Calculate max profit for rod length 4 by checking all cuts:
   - Cut 1: price[0] + rodCutting(3) = 1 + 8 = 9
   - Cut 2: price[1] + rodCutting(2) = 5 + 5 = 10
   - Cut 3: price[2] + rodCutting(1) = 8 + 1 = 9
   - Cut 4: price[3] + rodCutting(0) = 9 + 0 = 9

2. Step 2: Calculate rodCutting(3), rodCutting(2), rodCutting(1)

   rodCutting(3) = max(1+5=6, 5+1=6, 8+0=8) = 8
   rodCutting(2) = max(1+1=2, 5+0=5) = 5
   rodCutting(1) = 1

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Check all cuts and pick max sum recursively [OK]

Common Mistakes:

• Ignoring recursive calls
• Using wrong indices for prices
• Assuming max is price of full rod only

4. Identify the error in the following rod cutting function and select the fix:

function rodCutting(prices, n) {
  if (n <= 0) return 0;
  let maxVal = 0;
  for (let i = 1; i < n; i++) {
    maxVal = Math.max(maxVal, prices[i - 1] + rodCutting(prices, n - i));
  }
  return maxVal;
}

medium

A. Change loop condition to i <= n

B. Initialize maxVal to -Infinity

C. Change base case to if (n === 0) return 1

D. Remove recursive call inside loop

Solution

1. Step 1: Analyze loop range

   The loop runs from i=1 to i < n, so it misses the case when i = n (full rod cut).

2. Step 2: Fix loop condition

   Changing to i <= n includes the full rod cut option, ensuring all possibilities are checked.

3. Final Answer:

   Change loop condition to i <= n -> Option A

4. Quick Check:

   Loop must include full length cut = Change loop condition to i <= n [OK]

Hint: Loop must cover full rod length cuts [OK]

Common Mistakes:

• Using i < n instead of i <= n
• Incorrect base case return value
• Removing recursion breaks logic

5. Given a rod of length 5 and prices array [2, 5, 7, 8, 10], what is the maximum profit using dynamic programming approach?

hard

A. 14

B. 17

C. 15

D. 12

Solution

1. Step 1: Use bottom-up DP to calculate max profit

   Calculate max profit for lengths 1 to 5:
   Length 1: max = 2
   Length 2: max = max(5, 2+2) = 5
   Length 3: max = max(7, 5+2, 2+5) = 7
   Length 4: max = max(8, 7+2, 5+5, 2+7) = 10
   Length 5: max = max(10, 8+2, 7+5, 5+7, 2+10) = 17

2. Step 2: Confirm maximum profit for length 5

   The best cut is length 1 + length 4 (2+15=17) or length 2 + length 3 (5+7=12), so 17 is maximum.

3. Final Answer:

   17 -> Option B

4. Quick Check:

   DP max profit for length 5 = 17 [OK]

Hint: Use DP table from length 1 to n for max profit [OK]

Common Mistakes:

• Not considering all cut combinations
• Using recursive without memoization causing slow results
• Confusing prices index with rod length