Fibonacci Using Recursion in DSA Typescript - Time & Space Complexity

Time Complexity: Fibonacci Using Recursion

O(2^n)

Understanding Time Complexity

We want to understand how long it takes to find a Fibonacci number using recursion.

Specifically, how the time grows as the input number gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code calculates the nth Fibonacci number by calling itself twice for smaller numbers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two recursive calls for each number greater than 1.
• How many times: Each call splits into two more calls until reaching base cases.

How Execution Grows With Input

Each number requires two smaller Fibonacci calls, so the total calls grow very fast.

Input Size (n)	Approx. Operations
10	About 177 calls
20	About 21,891 calls
30	About 2,692,537 calls

Pattern observation: The number of calls roughly doubles with each increase in n, growing exponentially.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed doubles with each increase in the input number, making it very slow for large n.

Common Mistake

[X] Wrong: "The recursive Fibonacci runs in linear time because it just counts down from n to 0."

[OK] Correct: Each call makes two more calls, so the total calls grow much faster than just counting down.

Interview Connect

Understanding this helps you see why naive recursion can be slow and why techniques like memoization improve performance.

Self-Check

"What if we add memoization to this recursive Fibonacci? How would the time complexity change?"

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