Closures in Javascript - Time & Space Complexity

Time Complexity: Closures

O(n)

Understanding Time Complexity

We want to see how the time it takes to run code with closures changes as the input grows.

How does using closures affect the number of steps the program does?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function makeCounter() {
  let count = 0;
  return function() {
    count += 1;
    return count;
  };
}

const counter = makeCounter();
for (let i = 0; i < n; i++) {
  counter();
}
    

This code creates a counter function using a closure and calls it n times to increase the count.

Identify Repeating Operations

• Primary operation: Calling the inner function that increments the count.
• How many times: The loop calls this function n times.

How Execution Grows With Input

Each call to the counter function does a simple addition and returns a value.

Input Size (n)	Approx. Operations
10	10 increments
100	100 increments
1000	1000 increments

Pattern observation: The number of steps grows directly with n, so doubling n doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line with the number of times we call the closure.

Common Mistake

[X] Wrong: "Closures always make code slower because they add extra layers."

[OK] Correct: The closure itself just keeps data safe; calling the inner function still takes constant time each call, so time grows normally with input size.

Interview Connect

Understanding how closures affect time helps you explain your code clearly and shows you know how functions work under the hood.

Self-Check

"What if the inner function called itself recursively inside the closure? How would the time complexity change?"

Practice

1. What is a closure in JavaScript?

easy

A. A function that has no parameters

B. A function that runs only once

C. A function that returns another function without variables

D. A function that remembers variables from its creation context

Solution

1. Step 1: Understand closure definition

   A closure is a function that keeps access to variables from the place where it was created, even after that place has finished running.

2. Step 2: Match definition to options

   A function that remembers variables from its creation context correctly describes this behavior, while others describe unrelated concepts.

3. Final Answer:

   A function that remembers variables from its creation context -> Option D

4. Quick Check:

   Closure = function remembering variables [OK]

Hint: Closures remember variables from where they were made [OK]

Common Mistakes:

• Thinking closures are functions without parameters
• Confusing closures with immediately invoked functions
• Believing closures only run once

2. Which of the following is the correct syntax to create a closure in JavaScript?

easy

A. function outer() { let x = 5; function inner() { return x; } }

B. function outer() { let x = 5; return x; }

C. function outer() { let x = 5; return function inner() { return x; } }

D. function outer() { return function() { let x = 5; } }

Solution

1. Step 1: Identify closure syntax

   A closure requires an inner function that accesses variables from the outer function and is returned so it can be used later.

2. Step 2: Check options for correct closure

   function outer() { let x = 5; return function inner() { return x; } } returns an inner function that uses outer variable x, forming a closure. Others either don't return the function or don't access outer variables correctly.

3. Final Answer:

   function outer() { let x = 5; return function inner() { return x; } } -> Option C

4. Quick Check:

   Return inner function accessing outer variable [OK]

Hint: Closure needs returned inner function accessing outer variable [OK]

Common Mistakes:

• Not returning the inner function
• Returning only the variable, not a function
• Defining inner function but not using outer variables

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

function makeCounter() {
  let count = 0;
  return function() {
    count += 1;
    return count;
  };
}
const counter = makeCounter();
console.log(counter());
console.log(counter());

medium

A. 1 and 2

B. 0 and 1

C. undefined and undefined

D. 1 and 1

Solution

1. Step 1: Understand closure behavior in makeCounter

   The inner function increments count and returns it. The variable count is remembered by the closure.

2. Step 2: Trace the calls to counter()

   First call: count = 0 + 1 = 1, returns 1. Second call: count = 1 + 1 = 2, returns 2.

3. Final Answer:

   1 and 2 -> Option A

4. Quick Check:

   Closure keeps count incrementing: 1, 2 [OK]

Hint: Closure keeps variable state between calls [OK]

Common Mistakes:

• Thinking count resets each call
• Expecting output 1 and 1
• Confusing closure with global variable

4. Find the error in this closure example:

function createFuncs() {
  let funcs = [];
  for (var i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}
const functions = createFuncs();
console.log(functions[0]());
console.log(functions[1]());

medium

A. All functions return 3 instead of 0, 1, 2

B. Syntax error due to missing semicolons

C. Functions return undefined

D. Functions throw a ReferenceError

Solution

1. Step 1: Identify loop variable scope

   The loop uses var i, which is function-scoped, so all inner functions share the same i.

2. Step 2: Understand closure captures final i value

   After loop ends, i is 3. All functions return 3, not 0, 1, 2 as expected.

3. Final Answer:

   All functions return 3 instead of 0, 1, 2 -> Option A

4. Quick Check:

   var in loop causes shared i = 3 [OK]

Hint: var in loop shares variable; use let for block scope [OK]

Common Mistakes:

• Assuming each function captures separate i
• Thinking semicolons cause error
• Expecting ReferenceError or undefined

5. How can you fix the following code so each function remembers its own value of i?

function createFuncs() {
  let funcs = [];
  for (var i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}

hard

A. Use var but call the function immediately

B. Change var to let in the for loop declaration

C. Return i directly instead of a function

D. Declare i as a global variable

Solution

1. Step 1: Understand block scope with let

   Using let in the for loop creates a new binding of i for each iteration, so closures capture different i values.

2. Step 2: Compare with var behavior

   var is function-scoped, so all closures share the same i. Changing to let fixes this by giving each iteration its own i.

3. Final Answer:

   Change var to let in the for loop declaration -> Option B

4. Quick Check:

   let in loop creates separate i per iteration [OK]

Hint: Use let in loops to create separate closure variables [OK]

Common Mistakes:

• Trying to fix by making i global
• Returning i instead of function
• Using var with immediate call but misunderstanding closures