Call stack behavior in C++ - Time & Space Complexity

Time Complexity: Call stack behavior

O(n)

Understanding Time Complexity

When a program runs functions, it uses a call stack to keep track of where it is. Understanding how this stack grows helps us see how time grows with input.

We want to know how the number of function calls affects the total work done.

Scenario Under Consideration

Analyze the time complexity of the following recursive function.

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

This function calculates the factorial of a number by calling itself with smaller values until it reaches 1.

Identify Repeating Operations

Look at what repeats as the function calls itself:

• Primary operation: The recursive call to factorial(n - 1).
• How many times: It repeats once for each number from n down to 1, so n times.

How Execution Grows With Input

As n grows, the number of calls grows the same way.

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

Pattern observation: The work grows directly with n. Double n, double the calls.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows in a straight line with the input size.

Common Mistake

[X] Wrong: "Recursive calls happen all at once, so time is constant."

[OK] Correct: Each call waits for the next one to finish, so calls add up one after another, making time grow with n.

Interview Connect

Understanding how the call stack grows helps you explain recursion clearly and shows you can think about how programs use memory and time together.

Self-Check

What if the function called itself twice each time (like in Fibonacci)? How would the time complexity change?

Practice

1. What does the call stack do when a function is called in C++?

easy

A. It adds a new frame for the function call on top of the stack.

B. It removes the last function call from the stack.

C. It clears all previous function calls.

D. It pauses the program execution permanently.

Solution

1. Step 1: Understand function call behavior

   When a function is called, the program needs to remember where to return after the function finishes.

2. Step 2: Role of the call stack

   The call stack adds a new frame on top to store this information and local variables.

3. Final Answer:

   It adds a new frame for the function call on top of the stack. -> Option A

4. Quick Check:

   Function call = add frame [OK]

Hint: Call means add frame; return means remove frame [OK]

Common Mistakes:

• Thinking the stack removes frames on function call
• Believing the stack clears all calls at once
• Confusing call stack with program pause

2. Which of the following C++ code snippets correctly shows a function call that affects the call stack?

easy

A. int main() { return 0; } void foo() { foo(); }

B. void foo() { main(); } int main() { return 0; }

C. void foo() {} int main() { foo(); return 0; }

D. int main() { int x = 5; }

Solution

1. Step 1: Identify valid function call from main

   void foo() {} int main() { foo(); return 0; } calls foo() from main, which adds a frame to the call stack.

2. Step 2: Check other options for correctness

   void foo() { main(); } int main() { return 0; } calls main() from foo(), which is invalid as main should not be called explicitly. int main() { return 0; } void foo() { foo(); } has foo calling itself but foo is not called from main, so no stack change from main. int main() { int x = 5; } has no function calls.

3. Final Answer:

   void foo() {} int main() { foo(); return 0; } -> Option C

4. Quick Check:

   Valid call from main = void foo() {} int main() { foo(); return 0; } [OK]

Hint: Look for main calling another function directly [OK]

Common Mistakes:

• Thinking main can be called like a normal function
• Ignoring function call order
• Confusing declaration with call

3. What is the output of this C++ program?

#include <iostream>

void print(int n) {
  if (n == 0) return;
  print(n - 1);
  std::cout << n << " ";
}

int main() {
  print(3);
  return 0;
}

medium

A. 3 2 1

B. 1 2 3

C. 0 1 2 3

D. 3 2 1 0

Solution

1. Step 1: Trace recursive calls

   print(3) calls print(2), which calls print(1), which calls print(0). At print(0), it returns immediately.

2. Step 2: Print after returning from recursion

   After print(0) returns, print(1) prints '1 ', then print(2) prints '2 ', then print(3) prints '3 '.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Recursion prints after calls = 1 2 3 [OK]

Hint: Print happens after recursion call returns [OK]

Common Mistakes:

• Printing before recursive call instead of after
• Including 0 in output incorrectly
• Confusing order of prints in recursion

4. The following code causes a runtime error. What is the main cause?

void recurse(int n) {
  if (n > 0) {
    recurse(n + 1);
  }
}

int main() {
  recurse(1);
  return 0;
}

medium

A. Stack overflow due to infinite recursion increasing n.

B. Syntax error because of missing return statement.

C. No error; program runs and exits normally.

D. Variable n is not initialized.

Solution

1. Step 1: Analyze recursion condition

   The function calls recurse(n + 1) if n > 0, so n increases each call without stopping.

2. Step 2: Understand call stack effect

   Because n never reaches 0 or less, recursion never stops, causing stack frames to pile up until overflow.

3. Final Answer:

   Stack overflow due to infinite recursion increasing n. -> Option A

4. Quick Check:

   Infinite recursion = stack overflow [OK]

Hint: Check if recursion moves toward base case [OK]

Common Mistakes:

• Assuming recursion stops without base case
• Confusing syntax error with runtime error
• Ignoring infinite growth of parameter

5. Given this code, what will be the final value of result?

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

int main() {
  int result = factorial(4);
  return 0;
}

hard

A. 1

B. 10

C. 4

D. 24

Solution

1. Step 1: Understand factorial recursion

   factorial(4) calls factorial(3), factorial(3) calls factorial(2), factorial(2) calls factorial(1), which returns 1.

2. Step 2: Calculate return values step-by-step

   factorial(2) = 2 * 1 = 2; factorial(3) = 3 * 2 = 6; factorial(4) = 4 * 6 = 24.

3. Final Answer:

   24 -> Option D

4. Quick Check:

   4! = 24 [OK]

Hint: Multiply down from n to 1 recursively [OK]

Common Mistakes:

• Stopping recursion too early
• Adding instead of multiplying
• Confusing base case return value