Fibonacci Using Recursion in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the base case for Fibonacci recursion.

DSA C

int fibonacci(int n) {
    if (n == [1]) {
        return n;
    }
    return fibonacci(n - 1) + fibonacci(n - 2);
}

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

A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 as the base case.

Not defining a base case causing infinite recursion.

The base case for Fibonacci recursion is when n is 0, return 0.

2fill in blank

medium

Complete the code to define the second base case for Fibonacci recursion.

DSA C

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

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

A2

B3

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1 for the second base case.

Missing the second base case causing incorrect results.

The second base case is when n is 1, return 1.

3fill in blank

hard

Fix the error in the recursive call to correctly calculate Fibonacci.

DSA C

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

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causing infinite recursion.

Using * or / which are incorrect for Fibonacci.

The recursive call should subtract 1 from n to get the previous Fibonacci number.

4fill in blank

hard

Fill both blanks to complete the main function that prints Fibonacci numbers up to n.

DSA C

#include <stdio.h>

int fibonacci(int n);

int main() {
    int n = 5;
    for (int i = [1]; i [2] n; i++) {
        printf("%d ", fibonacci(i));
    }
    return 0;
}

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

A0

B<=

C<

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 1 missing the first Fibonacci number.

Using <= causing one extra number to print.

The loop starts at 0 and runs while i is less than n to print Fibonacci numbers from 0 to n-1.

5fill in blank

hard

Fill all three blanks to complete the recursive Fibonacci function with correct base cases and recursive calls.

DSA C

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

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

A0

B1

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up base case values.

Using addition instead of subtraction in recursive calls.

The base cases are n == 0 returning 0, n == 1 returning 1, and recursive calls subtract 1 and 2 from n.

Practice

1. What does the Fibonacci sequence represent when calculated using recursion?

easy

A. Each number is the sum of the two preceding numbers

B. Each number is the product of the two preceding numbers

C. Each number is the difference of the two preceding numbers

D. Each number is the division of the two preceding numbers

Solution

1. Step 1: Understand Fibonacci sequence definition

   The Fibonacci sequence is defined such that each number is the sum of the two numbers before it.

2. Step 2: Relate to recursion

   Recursion calls the function for the two previous numbers and adds their results to get the current number.

3. Final Answer:

   Each number is the sum of the two preceding numbers -> Option A

4. Quick Check:

   Fibonacci = sum of two previous numbers [OK]

Hint: Fibonacci adds previous two numbers recursively [OK]

Common Mistakes:

• Thinking Fibonacci multiplies previous numbers
• Confusing subtraction or division with addition
• Assuming Fibonacci uses only one previous number

2. Which of the following is the correct base case for a recursive Fibonacci function in C?

easy

A. if (n < 2) return 2;

B. if (n == 0) return 0; else return 1;

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

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

Solution

1. Step 1: Identify base cases for Fibonacci

   Fibonacci sequence starts with 0 and 1 for n=0 and n=1 respectively, so base case returns n when n is 0 or 1.

2. Step 2: Match base case syntax

   The condition if (n <= 1) return n; correctly returns 0 for n=0 and 1 for n=1.

3. Final Answer:

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

4. Quick Check:

   Base case returns n for n=0 or 1 [OK]

Hint: Base case returns n when n is 0 or 1 [OK]

Common Mistakes:

• Returning 2 for base case
• Returning 0 for n=1 incorrectly
• Missing base case leading to infinite recursion

3. What is the output of the following C code snippet?

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

int main() {
  printf("%d", fib(4));
  return 0;
}

medium

A. 4

B. 3

C. 5

D. 7

Solution

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

   fib(4) = fib(3) + fib(2) - fib(3) = fib(2) + fib(1) = (fib(1)+fib(0)) + 1 = (1+0)+1=2 - fib(2) = fib(1) + fib(0) = 1 + 0 = 1 So fib(4) = 2 + 1 = 3

2. Step 2: Confirm output printed

   The printf prints the integer 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   fib(4) = 3 [OK]

Hint: Trace fib calls down to base cases [OK]

Common Mistakes:

• Confusing fib(4) with fib(5)
• Adding wrong fib values
• Forgetting base cases

4. Identify the error in this recursive Fibonacci function and select the fix:

int fib(int n) {
  if (n == 0) return 0;
  else return fib(n-1) + fib(n-2);
}

medium

A. Add base case for n == 1 to return 1

B. Change return 0 to return 1 for n == 0

C. Remove else and just return fib(n-1) + fib(n-2)

D. Change function to iterative instead of recursive

Solution

1. Step 1: Check base cases

   Function only returns 0 for n==0 but misses base case for n==1, causing infinite recursion for n=1.

2. Step 2: Fix by adding base case for n==1

   Adding if (n == 1) return 1; fixes recursion and returns correct Fibonacci values.

3. Final Answer:

   Add base case for n == 1 to return 1 -> Option A

4. Quick Check:

   Both n=0 and n=1 base cases needed [OK]

Hint: Always include base cases for n=0 and n=1 [OK]

Common Mistakes:

• Missing base case for n=1
• Changing base case values incorrectly
• Removing else causing syntax confusion

5. You want to compute the 6th Fibonacci number using recursion but want to avoid repeated calculations. Which approach improves efficiency?

hard

A. Call fib(n-1) twice to ensure correctness

B. Increase the recursion depth limit

C. Use a loop instead of recursion without storing values

D. Use memoization to store and reuse computed Fibonacci values

Solution

1. Step 1: Understand repeated calculations in recursion

   Recursive Fibonacci calls fib(n-1) and fib(n-2) repeatedly, causing many duplicate calls and inefficiency.

2. Step 2: Use memoization to store results

   Memoization saves computed Fibonacci values in a cache, so repeated calls return stored results instantly, improving speed.

3. Final Answer:

   Use memoization to store and reuse computed Fibonacci values -> Option D

4. Quick Check:

   Memoization avoids repeated work [OK]

Hint: Store results to avoid repeated recursive calls [OK]

Common Mistakes:

• Increasing recursion depth doesn't fix inefficiency
• Using loops without caching still repeats work
• Calling fib(n-1) twice wastes time