Array Insertion at Beginning in DSA Javascript - Time & Space Complexity

Time Complexity: Array Insertion at Beginning

O(n)

Understanding Time Complexity

We want to understand how long it takes to add an item at the start of an array.

How does the time needed change as the array grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

const arr = [1, 2, 3, 4, 5];
arr.unshift(0); // Insert 0 at the beginning
console.log(arr);
    

This code adds a new element at the start of the array, shifting all other elements right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Shifting all existing elements one position to the right.
• How many times: Once for each element in the array.

How Execution Grows With Input

As the array gets bigger, the number of elements to move grows too.

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

Pattern observation: The work grows directly with the number of elements.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the start grows linearly with the array size.

Common Mistake

[X] Wrong: "Adding at the beginning is as fast as adding at the end."

[OK] Correct: Adding at the start requires moving all elements, while adding at the end just adds one item.

Interview Connect

Knowing how array operations scale helps you choose the right data structure and explain your choices clearly.

Self-Check

"What if we used a linked list instead of an array? How would the time complexity change?"

Practice

1. What does the unshift() method do in a JavaScript array?

easy

A. Adds one or more elements to the beginning of the array

B. Removes the first element from the array

C. Adds one or more elements to the end of the array

D. Removes the last element from the array

Solution

1. Step 1: Understand the unshift() method

   The unshift() method adds elements to the start of an array, shifting existing elements to the right.

2. Step 2: Compare with other array methods

   push() adds to the end, shift() removes from the start, and pop() removes from the end.

3. Final Answer:

   Adds one or more elements to the beginning of the array -> Option A

4. Quick Check:

   unshift() adds at start = B [OK]

Hint: Remember: unshift = add at front [OK]

Common Mistakes:

• Confusing unshift() with push()
• Thinking unshift() removes elements
• Mixing unshift() with shift()

2. Which of the following is the correct syntax to insert the value 10 at the beginning of array arr?

easy

A. arr.pushFront(10);

B. arr.addFirst(10);

C. arr.unshift(10);

D. arr.insertAt(0, 10);

Solution

1. Step 1: Recall JavaScript array methods

   JavaScript arrays use unshift() to add elements at the start.

2. Step 2: Check each option

   Only arr.unshift(10); is valid JavaScript syntax for this operation.

3. Final Answer:

   arr.unshift(10); -> Option C

4. Quick Check:

   Correct syntax for start insertion = A [OK]

Hint: Use unshift() to add at start, push() for end [OK]

Common Mistakes:

• Using non-existent methods like addFirst or pushFront
• Confusing unshift() with push()
• Trying to use insertAt which is not a JS array method

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

let arr = [2, 3, 4];
arr.unshift(1);
console.log(arr);

medium

A. [1, 3, 4]

B. [1, 2, 3, 4]

C. [2, 3, 4, 1]

D. [2, 3, 4]

Solution

1. Step 1: Understand unshift effect on array

   The unshift(1) adds 1 at the start, shifting existing elements right.

2. Step 2: Trace the array after unshift

   Original array: [2, 3, 4]
   After unshift(1): [1, 2, 3, 4]

3. Final Answer:

   [1, 2, 3, 4] -> Option B

4. Quick Check:

   unshift adds at front = [1, 2, 3, 4] [OK]

Hint: unshift adds element at front, shifts others right [OK]

Common Mistakes:

• Thinking unshift adds at end
• Ignoring that unshift shifts elements
• Expecting original array unchanged

4. Identify the error in the following code that tries to insert 0 at the beginning of array nums:

let nums = [1, 2, 3];
nums.unshift = 0;
console.log(nums);

medium

A. Using assignment instead of calling unshift as a function

B. Array nums is not defined

C. unshift cannot add zero

D. Syntax error due to missing parentheses

Solution

1. Step 1: Analyze the assignment to unshift property

   The code assigns 0 to nums.unshift, overwriting the method instead of calling it.

2. Step 2: Correct usage of unshift

   To add 0 at start, call nums.unshift(0); as a function, not assign a value.

3. Final Answer:

   Using assignment instead of calling unshift as a function -> Option A

4. Quick Check:

   unshift must be called, not assigned = A [OK]

Hint: Call unshift() as function, don't assign it a value [OK]

Common Mistakes:

• Assigning value to unshift instead of calling it
• Missing parentheses when calling unshift
• Thinking unshift is a property, not a method

5. Given the array let data = [5, 6, 7];, which code snippet correctly inserts the numbers 2, 3, 4 at the beginning so that the final array is [2, 3, 4, 5, 6, 7]?

hard

A. data.unshift([2, 3, 4]);

B. data.splice(0, 0, [2, 3, 4]);

C. data.push(2, 3, 4);

D. data.unshift(2, 3, 4);

Solution

1. Step 1: Understand unshift with multiple arguments

   The unshift() method can take multiple arguments and adds them all at the start in order.

2. Step 2: Analyze each option's effect

   data.unshift([2, 3, 4]); adds a single array as first element, resulting in nested array.
   data.unshift(2, 3, 4); adds 2, 3, 4 individually at start, producing desired array.
   data.push(2, 3, 4); adds elements at end.
   data.splice(0, 0, [2, 3, 4]); inserts a single array at index 0, nesting again.

3. Final Answer:

   data.unshift(2, 3, 4); -> Option D

4. Quick Check:

   unshift multiple args adds all at front = D [OK]

Hint: Pass multiple values to unshift() to add all at start [OK]

Common Mistakes:

• Passing an array instead of separate values to unshift
• Using push() which adds at end
• Using splice() incorrectly causing nested arrays