Rod Cutting Problem in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum revenue array.

DSA Java

int[] maxRevenue = new int[[1] + 1];

Drag options to blanks, or click blank then click option'

An

Bn * 2

Cn - 1

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using array size n causes index out of bounds when accessing maxRevenue[n].

Using n-1 misses the last length.

We need an array of size n+1 to store maximum revenues for lengths from 0 to n.

2fill in blank

medium

Complete the code to set the base case revenue for rod length 0.

DSA Java

maxRevenue[0] = [1];

Drag options to blanks, or click blank then click option'

A0

Bn

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting base revenue to 1 or negative values causes incorrect results.

Revenue for rod length 0 is zero because no cuts mean no revenue.

3fill in blank

hard

Fix the error in the loop that calculates maximum revenue for each rod length i.

DSA Java

for (int i = 1; i <= n; i++) {
    int maxVal = Integer.MIN_VALUE;
    for (int j = 1; j <= i; j++) {
        maxVal = Math.max(maxVal, price[j - 1] + [1][i - j]);
    }
    maxRevenue[i] = maxVal;
}

Drag options to blanks, or click blank then click option'

AmaxRevenue

Bprice

Ci

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using price instead of maxRevenue for the remaining length causes wrong calculation.

We must add the revenue for the remaining rod length i-j, which is stored in maxRevenue array.

4fill in blank

hard

Fill both blanks to complete the method signature and return statement for the rod cutting solution.

DSA Java

public static int [1](int[] price, int [2]) {
    // method body
    return maxRevenue[n];
}

Drag options to blanks, or click blank then click option'

AcutRod

Blength

Cn

DmaxRevenue

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect method names or parameter names causes confusion.

The method name is cutRod and the rod length parameter is commonly named n.

5fill in blank

hard

Fill all three blanks to complete the code that prints the maximum revenue for a rod of length n.

DSA Java

int[] price = {1, 5, 8, 9, 10, 17, 17, 20};
int [1] = price.length;
int maxProfit = [2](price, [3]);
System.out.println("Maximum Obtainable Value is " + maxProfit);

Drag options to blanks, or click blank then click option'

An

BcutRod

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names or method names causes errors.

We define n as the length of price array, call cutRod with price and n, then print maxProfit.

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