Stack overflow concept in Javascript - Time & Space Complexity

Time Complexity: Stack overflow concept

O(n)

Understanding Time Complexity

When a program runs, it uses a special memory area called the call stack to keep track of function calls.

We want to understand how the number of function calls affects the program's running steps and memory use.

Scenario Under Consideration

Analyze the time complexity of the following recursive function that calls itself without a stopping condition.

function infiniteRecursion() {
  return infiniteRecursion();
}

infiniteRecursion();
    

This code calls the same function again and again without stopping, causing the call stack to grow endlessly.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive function calls without end.
• How many times: Infinite times until the program crashes.

How Execution Grows With Input

Each call adds a new layer to the call stack, growing step by step without stopping.

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

Pattern observation: The number of calls grows directly with input, but here input is the depth of calls, which never ends.

Final Time Complexity

Time Complexity: O(n)

This means the number of steps grows directly with the number of recursive calls before the program crashes.

Common Mistake

[X] Wrong: "The recursion will stop on its own without a base case."

[OK] Correct: Without a stopping point, the function keeps calling itself forever, causing the program to crash with a stack overflow.

Interview Connect

Understanding how recursive calls use the call stack helps you write safe functions and avoid crashes in real projects.

Self-Check

"What if we add a base case to stop the recursion after a certain number of calls? How would the time complexity change?"

Practice

1. What causes a stack overflow error in JavaScript?

easy

A. Declaring global variables

B. Using too many variables in a function

C. Too many nested function calls without stopping

D. Calling a function only once

Solution

1. Step 1: Understand what the call stack is

   The call stack keeps track of function calls in JavaScript. Each call uses some memory.

2. Step 2: Identify what causes overflow

   If functions call themselves too many times without stopping, the stack runs out of memory, causing overflow.

3. Final Answer:

   Too many nested function calls without stopping -> Option C

4. Quick Check:

   Stack overflow = too many nested calls [OK]

Hint: Stack overflow = infinite or too deep recursion [OK]

Common Mistakes:

• Thinking global variables cause stack overflow
• Confusing variable count with call stack size
• Assuming single function call causes overflow

2. Which of the following JavaScript function definitions will cause a stack overflow when called?

easy

A. function infinite() { infinite(); }

B. function countDown(n) { if(n <= 0) return; countDown(n - 1); }

C. function greet() { console.log('Hello'); }

D. function add(a, b) { return a + b; }

Solution

1. Step 1: Check each function for recursion

   function countDown(n) { if(n <= 0) return; countDown(n - 1); } has a base case stopping recursion. function infinite() { infinite(); } calls itself endlessly without a base case. Options C and D are normal functions without recursion.

2. Step 2: Identify which causes stack overflow

   Only function infinite() { infinite(); } causes infinite recursion, leading to stack overflow error.

3. Final Answer:

   function infinite() { infinite(); } -> Option A

4. Quick Check:

   Infinite recursion = stack overflow [OK]

Hint: Infinite recursion without base case causes overflow [OK]

Common Mistakes:

• Assuming all recursive functions cause overflow
• Ignoring base case in recursion
• Confusing normal functions with recursive ones

3. What will be the output of this code?

function recurse(n) {
  if (n === 0) return 'Done';
  return recurse(n - 1);
}
console.log(recurse(3));

medium

A. 3

B. Stack overflow error

C. undefined

D. "Done"

Solution

1. Step 1: Trace the recursive calls

   recurse(3) calls recurse(2), then recurse(1), then recurse(0). At n=0, it returns 'Done'.

2. Step 2: Return values propagate back

   Each call returns the result of the next call, so finally recurse(3) returns 'Done'.

3. Final Answer:

   "Done" -> Option D

4. Quick Check:

   Base case stops recursion, returns 'Done' [OK]

Hint: Base case returns value, recursion unwinds safely [OK]

Common Mistakes:

• Expecting stack overflow without base case
• Thinking recursion returns input number
• Ignoring return propagation

4. Find the error in this recursive function that causes a stack overflow:

function factorial(n) {
  return n * factorial(n - 1);
}

medium

A. Wrong multiplication operator

B. Missing base case to stop recursion

C. Function name is incorrect

D. Should use addition instead of multiplication

Solution

1. Step 1: Check for base case in recursion

   The function calls itself without any condition to stop, so recursion never ends.

2. Step 2: Understand stack overflow cause

   Without a base case, calls pile up until memory runs out, causing stack overflow.

3. Final Answer:

   Missing base case to stop recursion -> Option B

4. Quick Check:

   Base case missing = infinite recursion [OK]

Hint: Always add base case to recursive functions [OK]

Common Mistakes:

• Assuming multiplication causes error
• Ignoring need for base case
• Thinking function name affects recursion

5. You want to safely calculate the sum of numbers from 1 to n using recursion without causing stack overflow. Which change should you make to this function?

function sum(n) {
  return n + sum(n - 1);
}

hard

A. Add a base case: if (n <= 0) return 0;

B. Change + to * operator

C. Call sum(n + 1) instead

D. Remove the recursive call

Solution

1. Step 1: Identify missing base case

   The function calls itself with n-1 but never stops, causing overflow.

2. Step 2: Add base case to stop recursion

   Adding if (n <= 0) return 0; stops recursion when n reaches 0 or less.

3. Step 3: Confirm safe recursion

   With base case, recursion ends properly and sum is calculated without overflow.

4. Final Answer:

   Add a base case: if (n <= 0) return 0; -> Option A

5. Quick Check:

   Base case stops recursion safely [OK]

Hint: Base case prevents infinite recursion and overflow [OK]

Common Mistakes:

• Changing operator instead of adding base case
• Increasing n causing infinite recursion
• Removing recursion loses function purpose