Recursion Base Case and Recursive Case in DSA C++ - Time & Space Complexity

Time Complexity: Recursion Base Case and Recursive Case

O(n)

Understanding Time Complexity

We want to understand how the time taken by a recursive function grows as the input size increases.

Specifically, how the base case and recursive calls affect the total work done.

Scenario Under Consideration

Analyze the time complexity of the following recursive function.

int factorial(int n) {
  if (n == 0) return 1;  // base case
  return n * factorial(n - 1);  // recursive case
}
    

This function calculates the factorial of n by calling itself with smaller values until it reaches zero.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One recursive call per function call.
• How many times: The function calls itself n times until it reaches the base case.

How Execution Grows With Input

Each call does a small amount of work and then calls itself with n-1.

Input Size (n)	Approx. Operations
10	About 10 calls
100	About 100 calls
1000	About 1000 calls

Pattern observation: The number of calls grows directly with n, so the work grows linearly.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows in a straight line as the input number increases.

Common Mistake

[X] Wrong: "Recursion always takes exponential time because it calls itself multiple times."

[OK] Correct: This function calls itself only once per call, so the time grows linearly, not exponentially.

Interview Connect

Understanding how base and recursive cases affect time helps you explain and optimize recursive solutions clearly.

Self-Check

"What if the function made two recursive calls instead of one? How would the time complexity change?"

Practice

1. What is the purpose of the base case in a recursive function?

easy

A. To print the result of the recursion

B. To call the function again with a smaller input

C. To increase the input size for the next call

D. To stop the recursion and prevent infinite calls

Solution

1. Step 1: Understand recursion basics

   Recursion works by calling itself repeatedly until a stopping point is reached.

2. Step 2: Identify the role of base case

   The base case stops further recursive calls to avoid infinite loops.

3. Final Answer:

   To stop the recursion and prevent infinite calls -> Option D

4. Quick Check:

   Base case = stop recursion [OK]

Hint: Base case stops recursion to avoid infinite loops [OK]

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case calls recursion again
• Ignoring the need for a stopping condition

2. Which of the following is the correct syntax for a base case in this recursive function to calculate factorial?

int factorial(int n) {
  if (___) return 1;
  return n * factorial(n - 1);
}

easy

A. n > 0

B. n == 0

C. n == 1

D. n < 0

Solution

1. Step 1: Identify factorial base case

   Factorial of 0 is defined as 1, so base case is when n == 0.

2. Step 2: Check options for correct base case

   n == 0 matches n == 0, which returns 1 stopping recursion correctly.

3. Final Answer:

   n == 0 -> Option B

4. Quick Check:

   Factorial base case = n == 0 [OK]

Hint: Factorial base case is n equals zero [OK]

Common Mistakes:

• Using n > 0 as base case causes infinite recursion
• Using n < 0 which is invalid for factorial
• Confusing base case with recursive case

3. What will be the output of the following code?

int sum(int n) {
  if (n == 0) return 0;
  return n + sum(n - 1);
}

int main() {
  std::cout << sum(3);
  return 0;
}

medium

A. 0

B. 3

C. 6

D. Error

Solution

1. Step 1: Trace recursive calls for sum(3)

   sum(3) = 3 + sum(2), sum(2) = 2 + sum(1), sum(1) = 1 + sum(0), sum(0) = 0.

2. Step 2: Calculate total sum

   sum(3) = 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Sum 1 to 3 = 6 [OK]

Hint: Sum from n to 0 adds all numbers [OK]

Common Mistakes:

• Returning n instead of n + sum(n-1)
• Missing base case causing infinite recursion
• Confusing output with just n

4. Identify the error in this recursive function to find factorial:

int factorial(int n) {
  if (n == 1) return 1;
  else return n * factorial(n + 1);
}

medium

A. Recursive call increases n, causing infinite recursion

B. Function should return void, not int

C. Missing return statement in base case

D. Base case should be n == 0, not n == 1

Solution

1. Step 1: Analyze recursive call

   The function calls factorial(n + 1), which increases n each time, never reaching base case.

2. Step 2: Identify infinite recursion cause

   Because n grows, base case n == 1 is never met, causing infinite recursion.

3. Final Answer:

   Recursive call increases n, causing infinite recursion -> Option A

4. Quick Check:

   Recursive call must reduce n [OK]

Hint: Recursive calls must reduce input to reach base case [OK]

Common Mistakes:

• Increasing n instead of decreasing in recursion
• Wrong base case value
• Ignoring infinite recursion risk

5. You want to write a recursive function to compute the nth Fibonacci number where fib(0) = 0 and fib(1) = 1. Which of the following correctly defines the base cases and recursive case?

hard

A. if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

B. if (n == 1) return 0; if (n == 2) return 1; else return fib(n - 1) + fib(n - 2);

C. if (n <= 1) return n; else return fib(n + 1) + fib(n - 2);

D. if (n == 0) return 1; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2);

Solution

1. Step 1: Identify correct base cases for Fibonacci

   By definition, fib(0) = 0 and fib(1) = 1 are base cases.

2. Step 2: Check recursive case correctness

   Recursive case sums fib(n-1) and fib(n-2), matching Fibonacci definition.

3. Final Answer:

   if (n == 0) return 0; if (n == 1) return 1; else return fib(n - 1) + fib(n - 2); -> Option A

4. Quick Check:

   Fibonacci base and recursive cases match definition [OK]

Hint: Base cases fib(0)=0, fib(1)=1; recursive sums previous two [OK]

Common Mistakes:

• Wrong base case values
• Increasing n in recursive calls
• Returning wrong values for base cases