Fibonacci Using Recursion in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case value for Fibonacci of 0.

DSA Javascript

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

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

A0

B1

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for n=0.

Returning n which is unnecessary here.

The Fibonacci of 0 is defined as 0, so the function should return 0 for n === 0.

2fill in blank

medium

Complete the code to return the base case value for Fibonacci of 1.

DSA Javascript

function fibonacci(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'

A-1

B0

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for n=1.

Returning n which is unnecessary here.

The Fibonacci of 1 is defined as 1, so the function should return 1 for n === 1.

3fill in blank

hard

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

DSA Javascript

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

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

An - 1

Bn + 1

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 which causes infinite recursion.

Using n which causes infinite recursion.

The Fibonacci sequence is calculated by adding fibonacci(n-1) and fibonacci(n-2). The first recursive call must be fibonacci(n - 1).

4fill in blank

hard

Fill both blanks to complete the recursive Fibonacci function with base cases.

DSA Javascript

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

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

A0

B1

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the base cases.

Using 2 or -1 as base cases.

The base cases for Fibonacci are n=0 returning 0 and n=1 returning 1.

5fill in blank

hard

Fill all three blanks to create a recursive Fibonacci function that returns the nth Fibonacci number.

DSA Javascript

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

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

An

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for parameter and conditions.

Mixing up base case values.

The function parameter is n, the first if checks if n === 0, and the second if checks if n === 1.

Practice

1. What does a recursive Fibonacci function do when called with fib(4)?

easy

A. Throws a syntax error because recursion is not allowed

B. Returns 4 immediately without any calculation

C. Runs an infinite loop without returning

D. Calculates the 4th Fibonacci number by calling itself with smaller numbers

Solution

1. Step 1: Understand recursive calls for fib(4)

   The function calls fib(3) and fib(2) to calculate fib(4).

2. Step 2: Combine results of smaller calls

   fib(3) and fib(2) are calculated recursively and their sum is returned as fib(4).

3. Final Answer:

   Calculates the 4th Fibonacci number by calling itself with smaller numbers -> Option D

4. Quick Check:

   Recursive calls break problem into smaller parts [OK]

Hint: Recursion breaks problem into smaller calls [OK]

Common Mistakes:

• Thinking recursion returns input directly
• Assuming recursion causes infinite loop always
• Confusing recursion with iteration

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

easy

A. if (n = 0) return 0;

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

C. if n == 1 then return 1

D. if (n > 1) return n;

Solution

1. Step 1: Identify correct JavaScript syntax for base case

   The base case must check if n is 0 or 1 and return n using proper JavaScript syntax.

2. Step 2: Validate each option

   if (n <= 1) return n; uses correct syntax: if (n <= 1) return n; Options A and B have syntax errors or wrong operators, D is logically incorrect.

3. Final Answer:

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

4. Quick Check:

   Base case uses if (n <= 1) return n; [OK]

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

Common Mistakes:

• Using assignment (=) instead of comparison (== or ===)
• Missing parentheses in if condition
• Incorrect logical condition for base case

3. What is the output of the following code?

function fib(n) {
  if (n <= 1) return n;
  return fib(n - 1) + fib(n - 2);
}
console.log(fib(5));

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Trace fib(5) recursive calls

   fib(5) = fib(4) + fib(3), fib(4) = fib(3) + fib(2), fib(3) = fib(2) + fib(1), and so on until base cases.

2. Step 2: Calculate values bottom-up

   fib(0)=0, fib(1)=1, fib(2)=1, fib(3)=2, fib(4)=3, fib(5)=5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing index with Fibonacci value
• Miscounting recursive calls
• Assuming fib(5) is 8 or 13 incorrectly

4. Identify the error in this recursive Fibonacci function:

function fib(n) {
  if (n == 0) return 0;
  if (n == 1) return 1;
  return fib(n - 1) - fib(n - 2);
}

medium

A. The recursive step uses subtraction instead of addition

B. The base cases are incorrect

C. The function will cause infinite recursion

D. The function syntax is invalid

Solution

1. Step 1: Check base cases correctness

   Base cases for n==0 and n==1 correctly return 0 and 1 respectively.

2. Step 2: Analyze recursive step

   The recursive step uses subtraction (fib(n - 1) - fib(n - 2)) instead of addition, which is incorrect for Fibonacci.

3. Final Answer:

   The recursive step uses subtraction instead of addition -> Option A

4. Quick Check:

   Fibonacci sums previous two, not subtracts [OK]

Hint: Fibonacci adds previous two numbers, not subtracts [OK]

Common Mistakes:

• Using wrong operator (+ vs -)
• Confusing base cases
• Assuming syntax error when logic is wrong

5. How can you optimize the recursive Fibonacci function to avoid repeated calculations?

hard

A. Remove base cases to speed up recursion

B. Increase the recursion depth limit

C. Use memoization to store and reuse computed values

D. Replace recursion with a while loop only

Solution

1. Step 1: Understand repeated calculations in recursion

   Recursive Fibonacci recalculates same values many times, causing inefficiency.

2. Step 2: Apply memoization to store results

   Memoization saves results of fib(n) calls in a cache to reuse and avoid repeated work.

3. Final Answer:

   Use memoization to store and reuse computed values -> Option C

4. Quick Check:

   Memoization improves recursion efficiency [OK]

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

Common Mistakes:

• Thinking increasing recursion limit helps
• Believing loops are only solution
• Removing base cases causing infinite recursion