Rod Cutting Problem in DSA Java - 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

We try all possible first cuts, combine their prices with best solutions for the leftover rod, and pick the best total price.

Execution Sample

DSA Java

int cutRod(int[] prices, int n) {
  if (n == 0) return 0;
  int max_val = Integer.MIN_VALUE;
  for (int i = 1; i <= n; i++) {
    max_val = Math.max(max_val, prices[i-1] + cutRod(prices, n - i));
  }
  return max_val;
}

This code finds the best price for a rod of length n by trying all cuts and choosing the maximum.

Execution Table

Step	Operation	Rod Length (n)	Cut Length (i)	Price for cut i	Best price for remaining (n-i)	Max price so far	Visual State (DP array)
1	Start	4	-	-	-	-∞	[ - , - , - , - , - ]
2	Try cut i=1	4	1	2	cutRod(prices,3)	-∞	[ - , - , - , - , - ]
3	Try cut i=1 subcall	3	-	-	-	-	[ - , - , - , - , - ]
4	Try cut i=1	3	1	2	cutRod(prices,2)	-∞	[ - , - , - , - , - ]
5	Try cut i=1 subcall	2	-	-	-	-	[ - , - , - , - , - ]
6	Try cut i=1	2	1	2	cutRod(prices,1)	-∞	[ - , - , - , - , - ]
7	Try cut i=1 subcall	1	-	-	-	-	[ - , - , - , - , - ]
8	Try cut i=1	1	1	2	cutRod(prices,0)=0	max(-∞, 2+0)=2	[ - , 2 , - , - , - ]
9	Try cut i=1	2	1	2	2	max(-∞, 2+2)=4	[ - , 2 , 4 , - , - ]
10	Try cut i=2	2	2	5	cutRod(prices,0)=0	max(4, 5+0)=5	[ - , 2 , 5 , - , - ]
11	Try cut i=1	3	1	2	5	max(-∞, 2+5)=7	[ - , 2 , 5 , 7 , - ]
12	Try cut i=2	3	2	5	cutRod(prices,1)=2	max(7, 5+2)=7	[ - , 2 , 5 , 7 , - ]
13	Try cut i=3	3	3	7	cutRod(prices,0)=0	max(7, 7+0)=7	[ - , 2 , 5 , 7 , - ]
14	Try cut i=1	4	1	2	7	max(-∞, 2+7)=9	[ - , 2 , 5 , 7 , 9 ]
15	Try cut i=2	4	2	5	cutRod(prices,2)=5	max(9, 5+5)=10	[ - , 2 , 5 , 7 , 10 ]
16	Try cut i=3	4	3	7	cutRod(prices,1)=2	max(10, 7+2)=10	[ - , 2 , 5 , 7 , 10 ]
17	Try cut i=4	4	4	8	cutRod(prices,0)=0	max(10, 8+0)=10	[ - , 2 , 5 , 7 , 10 ]
18	Return max price for length 4	4	-	-	-	10	[ - , 2 , 5 , 7 , 10 ]

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

Variable Tracker

Variable	Start	After Step 8	After Step 10	After Step 13	After Step 17	Final
n	4	1	2	3	4	4
i	-	1	2	3	4	4
max_val	-∞	2	5	7	10	10
DP array (max prices for lengths)	[ - , - , - , - , - ]	[ - , 2 , - , - , - ]	[ - , 2 , 5 , - , - ]	[ - , 2 , 5 , 7 , - ]	[ - , 2 , 5 , 7 , 10 ]	[ - , 2 , 5 , 7 , 10 ]

Key Moments - 3 Insights

Why do we try all cut lengths from 1 to n instead of just one?

Because the best price might come from cutting the rod into smaller pieces, not just selling it whole. The execution_table rows 2-17 show trying each cut length and updating max price.

Why does the recursion stop when n == 0?

Because a rod of length 0 has no value. This is the base case in the code (step 8), returning 0 to end recursion.

How does the DP array help avoid repeated calculations?

In this example, DP array is shown as visual state but not used in code. In optimized versions, it stores max prices for lengths to reuse. Here, it shows max prices found so far at each step (rows 8,10,13,17).

Visual Quiz - 3 Questions

Look at the execution_table at step 10, what is the max price so far for rod length 2?

A 5

B 2

C 4

D -∞

💡 Hint

Check the 'Max price so far' column at step 10 in execution_table

At which step does the max price for rod length 3 become 7?

A Step 9

B Step 14

C Step 11

D Step 17

💡 Hint

Look at the 'Max price so far' column for rod length 3 in execution_table rows 11-13

If the price for length 2 was increased from 5 to 6, what would be the max price for rod length 4 at step 17?

A 10

B 11

C 12

D 9

💡 Hint

Check how max price is calculated as max of cuts; increasing price for length 2 affects steps 15 and 17

Concept Snapshot

Rod Cutting Problem:
- Given rod length n and prices for each length
- Try all cuts i=1 to n
- Max price = max(prices[i-1] + best price for n-i)
- Use recursion or DP to avoid repeated work
- Goal: maximize total price by cutting rod optimally

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to maximize price. We try all cut lengths from 1 to n. For each cut, we add the price of that piece to the best price for the leftover rod. We keep track of the maximum price found. The recursion stops when rod length is zero. The execution table shows each step trying cuts and updating max price. The DP array stores best prices for smaller lengths to avoid repeated calculations. Key moments include understanding why all cuts are tried, the base case for recursion, and the role of DP. The visual quiz tests understanding of max price updates and effects of price changes.

Practice

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

easy

A. To find the shortest rod length possible

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

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

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

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 optimization 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 B

4. Quick Check:

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

Hint: Focus on maximizing profit, not minimizing cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing with counting ways to cut
• Assuming shortest rod length is the goal

2. Which of the following is the correct way to declare the price array for rod lengths in Java?

easy

A. int[] price = {1, 5, 8, 9};

B. int price[] = (1, 5, 8, 9);

C. int price = {1, 5, 8, 9};

D. int[] price = (1, 5, 8, 9);

Solution

1. Step 1: Recall Java array declaration syntax

   In Java, arrays are declared with square brackets and initialized with curly braces.

2. Step 2: Check each option

   int[] price = {1, 5, 8, 9}; uses correct syntax: int[] price = {1, 5, 8, 9};. Others use parentheses or wrong assignment.

3. Final Answer:

   int[] price = {1, 5, 8, 9}; -> Option A

4. Quick Check:

   Correct Java array syntax = int[] price = {1, 5, 8, 9}; [OK]

Hint: Use curly braces {} to initialize arrays in Java [OK]

Common Mistakes:

• Using parentheses () instead of curly braces {}
• Missing square brackets [] in declaration
• Assigning array to int variable

3. What will be the output of the following Java code snippet for rod length 4?

int[] price = {1, 5, 8, 9};
int n = 4;
int maxVal = 0;
for (int i = 1; i <= n; i++) {
    maxVal = Math.max(maxVal, price[i-1] + price[n - i]);
}
System.out.println(maxVal);

medium

A. 13

B. 9

C. 10

D. 8

Solution

1. Step 1: Trace the loop iterations and calculations

   For n=4, loop i=1 to 4:
   i=1: maxVal = max(0, price[0]+price[3]) = max(0,1+9)=10
   i=2: maxVal = max(10, price[1]+price[2])=max(10,5+8)=13
   i=3: maxVal = max(13, price[2]+price[1])=max(13,8+5)=13
   i=4: maxVal = max(13, price[3]+price[0])=max(13,9+1)=13

2. Step 2: Identify the final maxVal value

   After loop, maxVal is 13, but code prints maxVal after loop ends. The code prints 13.

3. Final Answer:

   13 -> Option A

4. Quick Check:

   Max profit for length 4 = 13 [OK]

Hint: Check all splits and sum prices to find max [OK]

Common Mistakes:

• Not considering all splits i=1 to n
• Off-by-one errors in array indexing
• Printing maxVal before loop ends

4. Identify the error in this Java method for rod cutting:

public int cutRod(int[] price, int n) {
    if (n <= 0) return 0;
    int maxVal = Integer.MIN_VALUE;
    for (int i = 1; i < n; i++) {
        maxVal = Math.max(maxVal, price[i-1] + cutRod(price, n - i));
    }
    return maxVal;
}

medium

A. Recursive call parameters are incorrect

B. Base case should return 1 instead of 0

C. Loop should run from i = 1 to i <= n

D. maxVal should be initialized to 0 instead of Integer.MIN_VALUE

Solution

1. Step 1: Analyze the loop range

   The loop runs from i=1 to i<n, so it misses the case i=n which is needed to consider the whole rod as one piece.

2. Step 2: Correct the loop condition

   The loop should run from i=1 to i<=n to include all possible cuts including the full length.

3. Final Answer:

   Loop should run from i = 1 to i <= n -> Option C

4. Quick Check:

   Include full length cut by looping to i <= n [OK]

Hint: Loop must include full rod length i = n [OK]

Common Mistakes:

• Using i < n instead of i <= n
• Incorrect base case return value
• Wrong initialization of maxVal

5. Given a rod of length 5 and price array int[] price = {2, 5, 7, 8, 10};, what is the maximum profit using dynamic programming approach?

hard

A. 14

B. 10

C. 15

D. 12

Solution

1. Step 1: Calculate max profit for each length up to 5

   Length 1: max = 2
   Length 2: max = max(5, 2+2=4) = 5
   Length 3: max = max(7, 5+2=7, 2+5=7) = 7
   Length 4: max = max(8, 7+2=9, 5+5=10, 2+7=9) = 10
   Length 5: max = max(10, 10+2=12, 7+5=12, 5+7=12, 2+10=12) = 12

2. Step 2: Check if cutting into smaller pieces yields better profit

   Try cutting length 5 into 1+4, 2+3, 3+2, 4+1 etc. The best is 2+3 or 3+2 or 1+4 giving 12, but also check if full length price 10 is less than these.

3. Step 3: Recalculate carefully

   Check all splits:
   1+4: 2 + 10 = 12
   2+3: 5 + 7 = 12
   3+2: 7 + 5 = 12
   4+1: 8 + 2 = 10
   But also check if cutting into 1+1+3 or 1+1+1+2 etc. can yield more.
   Try 1+1+3: 2+2+7=11
   Try 1+1+1+2: 2+2+2+5=11
   Try 1+1+1+1+1: 2*5=10
   So max is 12.

4. Final Answer:

   12 -> Option D

5. Quick Check:

   Max profit for length 5 = 12 [OK]

Hint: Check all splits and pick max sum for length 5 [OK]

Common Mistakes:

• Choosing price of full length only
• Not checking all split combinations
• Confusing price array indices