Fresh object literals vs variable assignment behavior in Typescript - Performance Comparison

Time Complexity: Fresh object literals vs variable assignment behavior

O(n)

Understanding Time Complexity

Let's explore how creating new objects or assigning variables affects the time it takes for a program to run.

We want to know how the number of operations changes when we use fresh object literals versus reusing variables.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function createObjects(n: number) {
  const arr = [];
  for (let i = 0; i < n; i++) {
    const obj = { value: i };
    arr.push(obj);
  }
  return arr;
}
    

This code creates an array of n fresh objects, each with a property value set to the loop index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating a new object literal inside the loop.
• How many times: Exactly n times, once per loop iteration.

How Execution Grows With Input

Each time we increase n, the loop runs more times, creating more objects.

Input Size (n)	Approx. Operations
10	10 object creations
100	100 object creations
1000	1000 object creations

Pattern observation: The number of operations grows directly with n. Double the input, double the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line as the input size increases.

Common Mistake

[X] Wrong: "Assigning the same object variable multiple times is just as slow as creating new objects each time."

[OK] Correct: Reusing the same object variable does not create new objects, so it does less work. Creating fresh objects inside a loop takes more time because each object is new and separate.

Interview Connect

Understanding how object creation inside loops affects performance helps you write clearer and more efficient code, a skill valued in many coding challenges and real projects.

Self-Check

What if we moved the object literal outside the loop and reused it? How would the time complexity change?

Practice

1. In TypeScript, what happens when you assign an object variable to another variable using let b = a;?

easy

A. The original object is deleted.

B. A new copy of the object is created.

C. The object is converted to a string.

D. Both variables point to the same object in memory.

Solution

1. Step 1: Understand variable assignment with objects

   Assigning one object variable to another copies the reference, not the object itself.

2. Step 2: Effect of reference copying

   Both variables now point to the same object in memory, so changes via one affect the other.

3. Final Answer:

   Both variables point to the same object in memory. -> Option D

4. Quick Check:

   Object assignment copies reference = A [OK]

Hint: Assigning objects copies references, not objects themselves [OK]

Common Mistakes:

• Thinking a new object is created on assignment
• Assuming variables hold independent copies
• Confusing object assignment with primitive assignment

2. Which of the following is the correct way to create a fresh object literal in TypeScript?

easy

A. let obj = obj2;

B. let obj = Object.assign({});

C. let obj = { name: 'Alice' };

D. let obj = new Object(obj2);

Solution

1. Step 1: Identify fresh object literal syntax

   Using curly braces {} creates a new, fresh object literal.

2. Step 2: Check other options

   let obj = Object.assign({}); is incorrect because Object.assign({}) needs a source object to copy from; let obj = obj2; copies reference; let obj = new Object(obj2); creates an object but with a reference or unexpected behavior.

3. Final Answer:

   let obj = { name: 'Alice' }; -> Option C

4. Quick Check:

   Fresh object literal uses {} = A [OK]

Hint: Use curly braces {} to create fresh objects [OK]

Common Mistakes:

• Using assignment to copy objects instead of fresh literals
• Misusing Object.assign without source
• Confusing new Object() with fresh literal

3. What will be the output of this TypeScript code?

let a = { x: 1 };
let b = a;
b.x = 5;
console.log(a.x);

medium

A. 1

B. 5

C. undefined

D. Error

Solution

1. Step 1: Understand reference assignment

   Variable b points to the same object as a.

2. Step 2: Modify property via b

   Changing b.x to 5 changes the shared object property.

3. Step 3: Log a.x

   Since a and b reference the same object, a.x is now 5.

4. Final Answer:

   5 -> Option B

5. Quick Check:

   Shared reference means change visible = B [OK]

Hint: Changing one reference changes all pointing to same object [OK]

Common Mistakes:

• Expecting original object to remain unchanged
• Confusing object copying with reference assignment
• Assuming primitive copy behavior for objects

4. Identify the bug in this TypeScript code snippet:

let obj1 = { val: 10 };
let obj2 = obj1;
obj2 = { val: 20 };
console.log(obj1.val);

medium

A. obj2 assignment creates a new object, obj1 remains unchanged.

B. Both obj1 and obj2 point to the new object.

C. Syntax error due to reassignment.

D. obj1.val will be 20, which is unexpected.

Solution

1. Step 1: Analyze initial assignment

   obj2 initially points to the same object as obj1.

2. Step 2: Reassign obj2 to a new object

   obj2 now points to a fresh object; obj1 still points to the original.

3. Step 3: Check console output

   obj1.val remains 10 because obj1's reference did not change.

4. Final Answer:

   obj2 assignment creates a new object, obj1 remains unchanged. -> Option A

5. Quick Check:

   Reassignment changes reference, original unaffected = D [OK]

Hint: Reassigning variable changes reference, original object stays same [OK]

Common Mistakes:

• Thinking obj1 changes when obj2 is reassigned
• Confusing property change with variable reassignment
• Expecting both variables to point to new object

5. You want to create two independent objects in TypeScript so that changing one does not affect the other. Which code snippet achieves this correctly?

const obj1 = { a: 1 };
const obj2 = ???
obj2.a = 5;
console.log(obj1.a);

hard

A. obj2 = { ...obj1 };

B. obj2 = Object.assign(obj1);

C. obj2 = obj1;

D. obj2 = obj1 as any;

Solution

1. Step 1: Understand shallow copy with spread operator

   Using { ...obj1 } creates a fresh object with copied properties.

2. Step 2: Check other options

   obj2 = obj1; copies reference; obj2 = Object.assign(obj1); is incorrect usage of Object.assign; obj2 = obj1 as any; is just a type assertion, no copy.

3. Step 3: Confirm independence

   Changing obj2.a does not affect obj1.a.

4. Final Answer:

   obj2 = { ...obj1 }; -> Option A

5. Quick Check:

   Spread operator creates fresh object copy = C [OK]

Hint: Use spread {...obj} to copy objects independently [OK]

Common Mistakes:

• Assigning reference instead of copying
• Misusing Object.assign without target
• Confusing type assertion with copying