Rod Cutting Problem in DSA Javascript - Execution Trace

Concept Flow - Rod Cutting Problem

Start with rod length n

↓

Try all cut lengths i from 1 to n

↓

Calculate price for cut i + best price for remaining length n-i

↓

Keep max price found

↓

Store max price for length n

↓

Repeat for smaller lengths

↓

Return max price for full rod length

We try all possible first cuts, combine their prices with the best prices for the leftover rod, and keep track of the maximum price for each rod length.

Execution Sample

DSA Javascript

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

This code finds the best price for a rod of length n by trying all cuts and recursively combining prices.

Execution Table

Step	Operation	Rod Length (n)	Cut Length (i)	Price for Cut (prices[i])	Remaining Length (n-i)	Best Price for Remaining	Max Price So Far	Visual State (DP array)
1	Start rodCut	4	-	-	-	-	0	[0,0,0,0,0]
2	Try cut i=1	4	1	1	3	?	?	[0,0,0,0,0]
3	Compute rodCut(3)	3	-	-	-	-	0	[0,0,0,0,0]
4	Try cut i=1	3	1	1	2	?	?	[0,0,0,0,0]
5	Compute rodCut(2)	2	-	-	-	-	0	[0,0,0,0,0]
6	Try cut i=1	2	1	1	1	?	?	[0,0,0,0,0]
7	Compute rodCut(1)	1	-	-	-	-	0	[0,0,0,0,0]
8	Try cut i=1	1	1	1	0	0	1	[0,0,0,0,0]
9	rodCut(1) returns	1	-	-	-	-	1	[0,1,0,0,0]
10	Try cut i=2	2	2	5	0	0	5	[0,1,0,0,0]
11	rodCut(2) returns	2	-	-	-	-	5	[0,1,5,0,0]
12	Try cut i=3	3	3	8	0	0	8	[0,1,5,0,0]
13	Try cut i=2	3	2	5	1	1	8	[0,1,5,8,0]
14	Try cut i=1	3	1	1	2	5	8	[0,1,5,8,0]
15	rodCut(3) returns	3	-	-	-	-	8	[0,1,5,8,0]
16	Try cut i=4	4	4	9	0	0	9	[0,1,5,8,0]
17	Try cut i=3	4	3	8	1	1	9	[0,1,5,8,9]
18	Try cut i=2	4	2	5	2	5	10	[0,1,5,8,9]
19	Try cut i=1	4	1	1	3	8	10	[0,1,5,8,10]
20	rodCut(4) returns	4	-	-	-	-	10	[0,1,5,8,10]

💡 All cuts tried for rod length 4, maximum price found is 10

Variable Tracker

Variable	Start	After Step 8	After Step 11	After Step 15	After Step 20
n	4	1	2	3	4
i	-	1	2	3	4
maxVal	0	1	5	8	10
DP array	[0,0,0,0,0]	[0,1,0,0,0]	[0,1,5,0,0]	[0,1,5,8,0]	[0,1,5,8,10]

Key Moments - 3 Insights

Why do we need to try all cut lengths i from 1 to n?

Because the best price might come from cutting the rod into smaller pieces, not just selling it whole. The execution_table rows 2-19 show trying all cuts to find the maximum.

Why does rodCut(0) return 0?

Because a rod of length 0 has no value. This base case stops recursion, as seen in step 8 where remaining length is 0 and price is 0.

How does the DP array help avoid repeated calculations?

In this example, the DP array is shown being filled after each return (steps 9, 11, 15, 20). Storing results prevents recalculating prices for the same lengths.

Visual Quiz - 3 Questions

Look at the execution_table at step 18, what is the max price so far for rod length 4?

A 8

B 9

C 10

D 5

💡 Hint

Check the 'Max Price So Far' column in row 18 of execution_table.

At which step does rodCut(2) return its final max price?

A Step 11

B Step 9

C Step 15

D Step 20

💡 Hint

Look for 'rodCut(2) returns' in the Operation column of execution_table.

If prices[2] changed from 5 to 7, how would maxVal at step 11 change?

A It would stay 5

B It would become 7

C It would become 8

D It would become 10

💡 Hint

At step 11, maxVal is prices[2] + rodCut(0). Changing prices[2] affects maxVal directly.

Concept Snapshot

Rod Cutting Problem:
- Goal: Maximize price by cutting rod into pieces
- Try all cuts i=1 to n
- maxVal = max(maxVal, price[i] + best for n-i)
- Use recursion or DP to store results
- Base case: rod length 0 returns 0

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to get the highest price. We try every possible first cut length, add its price to the best price for the leftover rod, and keep track of the maximum. The process repeats recursively until the rod length is zero, which returns zero price. Using a DP array helps store results to avoid repeated work. The execution table shows step-by-step how the function tries cuts, calculates prices, and updates the maximum price for each rod length.

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