Recursion Base Case and Recursive Case in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return 1 when n is 0, the base case for factorial.

DSA C++

int factorial(int n) {
    if (n == [1]) {
        return 1;
    }
    return n * factorial(n - 1);
}

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

A0

B-1

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 as base case

No base case causing infinite recursion

The base case for factorial is when n is 0, returning 1.

2fill in blank

medium

Complete the recursive call to compute factorial of n-1.

DSA C++

int factorial(int n) {
    if (n == 0) {
        return 1;
    }
    return n * factorial([1]);
}

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

An + 1

B0

Cn - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n+1 causing infinite recursion

Using 0 or 1 incorrectly

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

3fill in blank

hard

Fix the error in the base case condition to correctly stop recursion.

DSA C++

int sumToN(int n) {
    if (n [1] 0) {
        return 0;
    }
    return n + sumToN(n - 1);
}

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

A>

B==

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or > causing incorrect stopping

Using != causing infinite recursion

The base case should check if n equals 0 to stop recursion.

4fill in blank

hard

Fill both blanks to create a recursive function that sums numbers from 1 to n.

DSA C++

int sumToN(int n) {
    if (n [1] 0) {
        return 0;
    }
    return n [2] sumToN(n - 1);
}

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

A==

B+

C-

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operator

Using subtraction instead of addition

The base case checks if n == 0, and the recursive case adds n to sumToN(n-1).

5fill in blank

hard

Fill all three blanks to complete a recursive function that computes the nth Fibonacci number.

DSA C++

int fibonacci(int n) {
    if (n [1] 0) {
        return 0;
    } else if (n [2] 1) {
        return 1;
    }
    return fibonacci(n - 1) [3] fibonacci(n - 2);
}

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

A==

B+

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators

Using subtraction or other operators instead of addition

Base cases check if n == 0 or n == 1, and recursive case adds fibonacci(n-1) and fibonacci(n-2).

Practice

1. What is the purpose of the base case in a recursive function?

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 recursion basics

   Recursion works by calling itself repeatedly until a stopping point is reached.

2. Step 2: Identify the role of base case

   The base case stops further recursive calls to avoid infinite loops.

3. Final Answer:

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

4. Quick Check:

   Base case = stop recursion [OK]

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

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case calls recursion again
• Ignoring the need for a stopping condition

2. Which of the following is the correct syntax for a base case in this recursive function to calculate factorial?

int factorial(int n) {
  if (___) return 1;
  return n * factorial(n - 1);
}

easy

A. n > 0

B. n == 0

C. n == 1

D. n < 0

Solution

1. Step 1: Identify factorial base case

   Factorial of 0 is defined as 1, so base case is when n == 0.

2. Step 2: Check options for correct base case

   n == 0 matches n == 0, which returns 1 stopping recursion correctly.

3. Final Answer:

   n == 0 -> Option B

4. Quick Check:

   Factorial base case = n == 0 [OK]

Hint: Factorial base case is n equals zero [OK]

Common Mistakes:

• Using n > 0 as base case causes infinite recursion
• Using n < 0 which is invalid for factorial
• Confusing base case with recursive case

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

int sum(int n) {
  if (n == 0) return 0;
  return n + sum(n - 1);
}

int main() {
  std::cout << sum(3);
  return 0;
}

medium

A. 0

B. 3

C. 6

D. Error

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 C

4. Quick Check:

   Sum 1 to 3 = 6 [OK]

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

Common Mistakes:

• Returning n instead of n + sum(n-1)
• Missing base case causing infinite recursion
• Confusing output with just n

4. Identify the error in this recursive function to find factorial:

int factorial(int n) {
  if (n == 1) return 1;
  else return n * factorial(n + 1);
}

medium

A. Recursive call increases n, causing infinite recursion

B. Function should return void, not int

C. Missing return statement in base case

D. Base case should be n == 0, not n == 1

Solution

1. Step 1: Analyze recursive call

   The function calls factorial(n + 1), which increases n each time, never reaching base case.

2. Step 2: Identify infinite recursion cause

   Because n grows, base case n == 1 is never met, causing infinite recursion.

3. Final Answer:

   Recursive call increases n, causing infinite recursion -> Option A

4. Quick Check:

   Recursive call must reduce n [OK]

Hint: Recursive calls must reduce input to reach base case [OK]

Common Mistakes:

• Increasing n instead of decreasing in recursion
• Wrong base case value
• Ignoring infinite recursion risk

5. You want to write a recursive function to compute the nth Fibonacci number where fib(0) = 0 and fib(1) = 1. Which of the following correctly defines the base cases and recursive case?

hard

A. if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

B. if (n == 1) return 0; if (n == 2) return 1; else return fib(n - 1) + fib(n - 2);

C. if (n <= 1) return n; else return fib(n + 1) + fib(n - 2);

D. if (n == 0) return 1; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

Solution

1. Step 1: Identify correct base cases for Fibonacci

   By definition, fib(0) = 0 and fib(1) = 1 are base cases.

2. Step 2: Check recursive case correctness

   Recursive case sums fib(n-1) and fib(n-2), matching Fibonacci definition.

3. Final Answer:

   if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2); -> Option A

4. Quick Check:

   Fibonacci base and recursive cases match definition [OK]

Hint: Base cases fib(0)=0, fib(1)=1; recursive sums previous two [OK]

Common Mistakes:

• Wrong base case values
• Increasing n in recursive calls
• Returning wrong values for base cases