Fibonacci Using Recursion in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the base case in a recursive Fibonacci function?

The base case returns the first two Fibonacci numbers directly, usually when n is 0 or 1, returning n itself to stop further recursion.

beginner

Explain how recursion works in calculating Fibonacci numbers.

Recursion breaks the problem into smaller parts by calling the Fibonacci function for (n-1) and (n-2), then adds their results to get the nth Fibonacci number.

intermediate

What is the time complexity of the naive recursive Fibonacci algorithm?

The time complexity is exponential, specifically O(2^n), because it recalculates the same Fibonacci numbers multiple times.

intermediate

Why can recursion be inefficient for Fibonacci calculation without optimization?

Because it repeats calculations for the same inputs many times, leading to a large number of function calls and slow performance.

beginner

Show the TypeScript function signature for a recursive Fibonacci function.

function fibonacci(n: number): number

What does the recursive Fibonacci function return when n is 0?

A 0

B 1

C n

D -1

The base case for n=0 returns 0 as the first Fibonacci number.

Which two previous Fibonacci numbers does the recursive function add to get the nth number?

A n and n-1

B n-1 and n-2

C n-2 and n-3

D n-3 and n-4

The function calls fibonacci(n-1) and fibonacci(n-2) and adds their results.

What is a major drawback of the naive recursive Fibonacci approach?

A Does not work for n=0

B Uses too much memory

C Exponential time complexity

D Returns wrong results

The naive recursion has exponential time complexity due to repeated calculations.

Which technique can improve recursive Fibonacci efficiency?

A Memoization

B Sorting

C Loop unrolling

D Binary search

Memoization stores results of previous calls to avoid repeated work.

What is the output of fibonacci(3) using recursion?

A 1

B 3

C 5

D 2

fibonacci(3) = fibonacci(2) + fibonacci(1) = 1 + 1 = 2.

Describe how a recursive function calculates the nth Fibonacci number.

Think about how the function calls itself with smaller inputs.

Explain why naive recursion for Fibonacci is inefficient and how to improve it.

Consider what happens when the function calls overlap.

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 square of the previous number

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: Recognize recursion use

   Recursion calls the function itself to calculate previous two numbers and sums them.

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 last two numbers, not multiply or subtract [OK]

Common Mistakes:

• Thinking Fibonacci multiplies previous numbers
• Confusing subtraction or squares with addition
• Assuming Fibonacci is a product sequence

2. Which of the following is the correct TypeScript function signature for a recursive Fibonacci function?

easy

A. function fib(n: string): string

B. function fib(n: boolean): boolean

C. function fib(n: number): number

D. function fib(): void

Solution

1. Step 1: Identify input and output types

   The Fibonacci function takes a number input and returns a number output.

2. Step 2: Match correct function signature

   Only the signature with number input and number output matches the Fibonacci function requirements.

3. Final Answer:

   function fib(n: number): number -> Option C

4. Quick Check:

   Input and output types must be number [OK]

Hint: Fibonacci input/output are numbers, not strings or booleans [OK]

Common Mistakes:

• Using wrong parameter types like string or boolean
• Missing return type or using void
• Confusing function signature syntax

3. What is the output of the following TypeScript code?

function fib(n: number): number {
  if (n <= 1) return n;
  return fib(n - 1) + fib(n - 2);
}
console.log(fib(4));

medium

A. 5

B. 7

C. 4

D. 3

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

   The console.log prints 5 as the result of fib(4).

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fib(4) = 5 [OK]

Hint: Trace fib calls step-by-step to sum previous two results [OK]

Common Mistakes:

• Confusing fib(4) with fib(5)
• Miscounting base cases fib(0) and fib(1)
• Adding wrong recursive results

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

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

medium

A. Change base case n == 0 to n <= 0

B. Change subtraction to addition in return statement

C. Remove base cases completely

D. Change function to return string instead of number

Solution

1. Step 1: Analyze return statement

   The function uses subtraction (fib(n - 1) - fib(n - 2)) which is incorrect for Fibonacci.

2. Step 2: Correct the operation

   Fibonacci requires addition of previous two numbers, so change '-' to '+'.

3. Final Answer:

   Change subtraction to addition in return statement -> Option B

4. Quick Check:

   Fibonacci sums previous two numbers [OK]

Hint: Fibonacci adds, never subtracts previous two results [OK]

Common Mistakes:

• Using subtraction instead of addition
• Removing base cases causing infinite recursion
• Changing return type unnecessarily

5. Given the recursive Fibonacci function below, what is the output of fib(0) and fib(1)? Why are these base cases important?

function fib(n: number): number {
  if (n <= 1) return n;
  return fib(n - 1) + fib(n - 2);
}

hard

A. fib(0) = 0, fib(1) = 1; base cases are optional

B. fib(0) = 1, fib(1) = 0; base cases speed up recursion

C. fib(0) = 1, fib(1) = 1; base cases prevent wrong output

D. fib(0) = 0, fib(1) = 1; base cases stop infinite recursion

Solution

1. Step 1: Evaluate base cases fib(0) and fib(1)

   fib(0) returns 0 and fib(1) returns 1 as per the condition n <= 1 returns n.

2. Step 2: Understand importance of base cases

   Base cases stop the recursion from continuing infinitely by providing direct answers for smallest inputs.

3. Final Answer:

   fib(0) = 0, fib(1) = 1; base cases stop infinite recursion -> Option D

4. Quick Check:

   Base cases return 0 and 1 to end recursion [OK]

Hint: Base cases return n for n ≤ 1 to stop recursion [OK]

Common Mistakes:

• Mixing up fib(0) and fib(1) values
• Thinking base cases are optional
• Ignoring recursion stopping condition