Recursion Base Case and Recursive Case in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the base case for the factorial function.

DSA Java

if (n == [1]) {
    return 1;
}

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

A1

B0

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as base case which is incorrect for factorial.

Missing the base case causing infinite recursion.

The base case for factorial is when n equals 1, returning 1 stops the recursion.

2fill in blank

medium

Complete the code to write the recursive call for factorial.

DSA Java

return n * factorial([1]);

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An - 1

Bn

Cn + 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n instead of n-1 causing infinite recursion.

Adding 1 instead of subtracting.

The recursive call reduces n by 1 each time to approach the base case.

3fill in blank

hard

Fix the error in the recursive factorial function by completing the base case condition.

DSA Java

if ([1] <= 1) {
    return 1;
}

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

An < 0

Bn

Cn == 0

Dn > 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect variable or condition causing wrong base case.

Checking for n > 1 which is wrong.

The base case checks if n is less than or equal to 1 to stop recursion.

4fill in blank

hard

Fill both blanks to complete the recursive Fibonacci function.

DSA Java

if (n == [1] || n == [2]) {
    return n;
}

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

A0

B1

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 and 2 as base cases which is incorrect.

Missing one of the base cases.

Fibonacci base cases are when n is 0 or 1, returning n directly.

5fill in blank

hard

Fill both blanks to complete the recursive Fibonacci return statement.

DSA Java

return fibonacci([1]) + fibonacci([2]);

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

An - 1

Bn - 2

Cn + 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n+1 or n instead of n-1 and n-2.

Adding wrong indices causing infinite recursion.

The recursive Fibonacci function returns the sum of fibonacci(n-1) and fibonacci(n-2).

Practice

1. In recursion, what is the purpose of the base case?

easy

A. To print the result of the recursion

B. To call the function again with a smaller input

C. To increase the input size for the next call

D. To stop the recursion and prevent infinite calls

Solution

1. Step 1: Understand the base case role

   The base case is the condition where recursion stops to avoid infinite calls.

2. Step 2: Differentiate base case from recursive case

   The recursive case calls the function again, but the base case ends recursion.

3. Final Answer:

   To stop the recursion and prevent infinite calls -> Option D

4. Quick Check:

   Base case stops recursion [OK]

Hint: Base case stops recursion to avoid infinite loops [OK]

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case calls function again
• Ignoring base case leads to infinite recursion

2. Which of the following is the correct syntax for a base case in a recursive Java method that calculates factorial?

easy

A. if (n > 0) return n * factorial(n - 1);

B. return n * factorial(n + 1);

C. if (n == 0) return 1;

D. if (n == 1) factorial(n);

Solution

1. Step 1: Identify base case condition for factorial

   Factorial base case is when n equals 0, return 1.

2. Step 2: Check syntax correctness

   if (n == 0) return 1; correctly returns 1 when n == 0, stopping recursion.

3. Final Answer:

   if (n == 0) return 1; -> Option C

4. Quick Check:

   Base case for factorial is n == 0 [OK]

Hint: Base case returns fixed value, usually simplest input [OK]

Common Mistakes:

• Using wrong condition for base case
• Calling factorial with n + 1 causing infinite recursion
• Not returning a value in base case

3. What is the output of this Java code?

public class Test {
  public static int sum(int n) {
    if (n == 0) return 0;
    return n + sum(n - 1);
  }
  public static void main(String[] args) {
    System.out.println(sum(3));
  }
}

medium

A. StackOverflowError

B. 6

C. 0

D. 3

Solution

1. Step 1: Trace recursive calls for sum(3)

   sum(3) = 3 + sum(2), sum(2) = 2 + sum(1), sum(1) = 1 + sum(0), sum(0) = 0.

2. Step 2: Calculate total sum

   sum(3) = 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Sum 1+2+3 = 6 [OK]

Hint: Sum from n down to 0 adds all numbers [OK]

Common Mistakes:

• Confusing base case return value
• Adding n - 1 instead of n
• Expecting output 3 instead of sum

4. Identify the error in this recursive Java method for factorial:

public static int factorial(int n) {
  if (n == 0) return 0;
  return n * factorial(n - 1);
}

medium

A. Base case returns 0 instead of 1

B. Recursive call uses n + 1 instead of n - 1

C. Missing return statement in recursive case

D. No base case defined

Solution

1. Step 1: Check base case return value

   Factorial base case should return 1 for n == 0, but code returns 0.

2. Step 2: Understand impact of wrong base case

   Returning 0 causes factorial to always be 0 after recursion.

3. Final Answer:

   Base case returns 0 instead of 1 -> Option A

4. Quick Check:

   Factorial base case must return 1 [OK]

Hint: Base case for factorial returns 1, not 0 [OK]

Common Mistakes:

• Returning 0 in base case
• Using wrong recursive call direction
• Forgetting return keyword

5. Given this recursive method to compute Fibonacci numbers:

public static int fib(int n) {
  if (n <= 1) return n;
  return fib(n - 1) + fib(n - 2);
}

What is the output of fib(4) and explain why the base case is important here?

hard

A. 3; base case stops infinite recursion by returning n when n ≤ 1

B. 5; base case returns 0 always

C. 7; base case calls fib(0) and fib(1) repeatedly

D. StackOverflowError; no base case defined

Solution

1. Step 1: Calculate fib(4) using recursion

   fib(4) = fib(3) + fib(2); fib(3) = fib(2) + fib(1); fib(2) = fib(1) + fib(0); fib(1)=1, fib(0)=0.

2. Step 2: Compute values step-by-step

   fib(2)=1+0=1; fib(3)=1+1=2; fib(4)=2+1=3. Base case returns n when n ≤ 1, stopping recursion.

3. Final Answer:

   3; base case stops infinite recursion by returning n when n ≤ 1 -> Option A

4. Quick Check:

   fib(4) = 3 with base case stopping recursion [OK]

Hint: Base case returns n when n ≤ 1 to stop recursion [OK]

Common Mistakes:

• Miscomputing fib(4) as 5 or 7
• Ignoring base case importance
• Assuming base case returns 0 always