Fibonacci Using Recursion in DSA C - Time & Space Complexity

Time Complexity: Fibonacci Using Recursion

O(2^n)

Understanding Time Complexity

We want to understand how the time needed to find a Fibonacci number grows as the input number gets bigger.

How does the number of steps change when we use recursion to find Fibonacci numbers?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int fibonacci(int n) {
    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 values until it reaches the base cases.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to fibonacci for n-1 and n-2.
• How many times: Each call generates two more calls until base cases, causing many repeated calculations.

How Execution Grows With Input

Each number requires calculating two smaller Fibonacci numbers, which themselves require two smaller ones, and so on.

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 grows very fast, roughly doubling with each increase in n.

Final Time Complexity

Time Complexity: O(2^n)

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

Common Mistake

[X] Wrong: "Since the function calls itself twice, the time complexity is just twice the input size, or O(n)."

[OK] Correct: Each call creates two more calls, leading to an exponential number of calls, not just a simple multiple.

Interview Connect

Understanding this recursive pattern helps you recognize when recursion can be inefficient and when to use better methods like memoization or iteration.

Self-Check

"What if we add a memory to store results of fibonacci calls (memoization)? 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 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