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DSA Javascriptprogramming~5 mins

Push Using Linked List Node in DSA Javascript - Time & Space Complexity

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Time Complexity: Push Using Linked List Node
O(1)
Understanding Time Complexity

We want to understand how long it takes to add a new item to the start of a linked list.

How does the time needed change as the list grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }

  push(value) {
    const newNode = new Node(value);
    newNode.next = this.head;
    this.head = newNode;
  }
}
    

This code adds a new node at the start of the linked list by adjusting pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Creating a new node and updating pointers.
  • How many times: Exactly once per push call, no loops or recursion involved.
How Execution Grows With Input

Adding a node always takes the same steps, no matter how big the list is.

Input Size (n)Approx. Operations
103 (create node, set next, update head)
1003 (same steps, no change)
10003 (still the same fixed steps)

Pattern observation: The number of steps stays the same regardless of list size.

Final Time Complexity

Time Complexity: O(1)

This means adding a node at the start takes a constant amount of time no matter how big the list is.

Common Mistake

[X] Wrong: "Adding a node takes longer as the list grows because we have to move through the list."

[OK] Correct: We only change the head pointer and the new node's next pointer; we never traverse the list, so time stays constant.

Interview Connect

Knowing that pushing to the front of a linked list is fast and constant time shows you understand efficient data structure operations.

Self-Check

"What if we changed push to add a node at the end of the linked list? How would the time complexity change?"

Practice

(1/5)
1. What does the push operation do in a linked list when using nodes?
easy
A. Removes the last node from the list
B. Adds a new node at the beginning of the list
C. Searches for a node with a specific value
D. Deletes the first node of the list

Solution

  1. Step 1: Understand push operation in linked list

    The push operation adds a new node to the start (head) of the linked list.
  2. Step 2: Compare with other operations

    Removing or searching nodes are different operations, not push.
  3. Final Answer:

    Adds a new node at the beginning of the list -> Option B
  4. Quick Check:

    Push = Add at head [OK]
Hint: Push always adds node at the start of the list [OK]
Common Mistakes:
  • Confusing push with pop or delete
  • Thinking push adds at the end
  • Mixing push with search operation
2. Which of the following is the correct syntax to create a new node and push it to the front of a linked list in JavaScript?
easy
A. newNode = Node(value); head = newNode.next;
B. head = new Node(value); head.next = newNode;
C. let newNode = new Node(value); newNode.next = head; head = newNode;
D. head.next = new Node(value); head = head.next;

Solution

  1. Step 1: Create new node and link it to current head

    We create a new node, set its next to current head, then update head to new node.
  2. Step 2: Check each option

    let newNode = new Node(value); newNode.next = head; head = newNode; correctly does this. Others either reverse assignments or miss new keyword.
  3. Final Answer:

    let newNode = new Node(value); newNode.next = head; head = newNode; -> Option C
  4. Quick Check:

    New node points to old head, head updated [OK]
Hint: New node's next points to old head, then update head [OK]
Common Mistakes:
  • Forgetting to use 'new' keyword
  • Assigning head before setting newNode.next
  • Mixing order of assignments
3. What will be the linked list after executing the following code?
class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

let head = null;

function push(value) {
  let newNode = new Node(value);
  newNode.next = head;
  head = newNode;
}

push(10);
push(20);
push(30);
medium
A. 10 -> 20 -> 30 -> null
B. null -> 10 -> 20 -> 30
C. 20 -> 30 -> 10 -> null
D. 30 -> 20 -> 10 -> null

Solution

  1. Step 1: Trace push calls and list updates

    First push(10): head = 10 -> null
    Second push(20): head = 20 -> 10 -> null
    Third push(30): head = 30 -> 20 -> 10 -> null
  2. Step 2: Confirm final linked list order

    New nodes are added at front, so list is 30 -> 20 -> 10 -> null.
  3. Final Answer:

    30 -> 20 -> 10 -> null -> Option D
  4. Quick Check:

    Last pushed is first in list [OK]
Hint: Last pushed node becomes new head [OK]
Common Mistakes:
  • Assuming push adds at end
  • Confusing order of nodes
  • Ignoring that head updates each push
4. Identify the error in the following push function for a linked list:
function push(value) {
  let newNode = Node(value);
  newNode.next = head;
  head = newNode;
}
medium
A. Missing 'new' keyword when creating newNode
B. Incorrect assignment of newNode.next
C. head should be assigned before newNode.next
D. Function should return the new head

Solution

  1. Step 1: Check node creation syntax

    Node is a class; must use 'new' keyword to create instance. Missing here.
  2. Step 2: Verify other assignments

    newNode.next = head and head = newNode are correct order and assignments.
  3. Final Answer:

    Missing 'new' keyword when creating newNode -> Option A
  4. Quick Check:

    Use 'new' to create class instance [OK]
Hint: Always use 'new' with class constructors [OK]
Common Mistakes:
  • Forgetting 'new' causes runtime error
  • Swapping assignment order
  • Expecting push to return head
5. Given a linked list with nodes 5 -> 15 -> 25 -> null, what will be the list after pushing a new node with value 10 using the push function?
function push(value) {
  let newNode = new Node(value);
  newNode.next = head;
  head = newNode;
}

Assume head points to the first node (5).
medium
A. 10 -> 5 -> 15 -> 25 -> null
B. 5 -> 15 -> 25 -> 10 -> null
C. 5 -> 10 -> 15 -> 25 -> null
D. 25 -> 15 -> 5 -> 10 -> null

Solution

  1. Step 1: Understand push adds node at front

    Push creates new node with value 10, sets its next to current head (5), then updates head to new node.
  2. Step 2: Visualize new list order

    New head is 10, followed by old list 5 -> 15 -> 25 -> null, so final list is 10 -> 5 -> 15 -> 25 -> null.
  3. Final Answer:

    10 -> 5 -> 15 -> 25 -> null -> Option A
  4. Quick Check:

    Push adds new node at front [OK]
Hint: Push inserts new node before current head [OK]
Common Mistakes:
  • Appending new node at end instead of front
  • Not updating head pointer
  • Mixing node order after push
Recommended

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10 questions~5 mineasy hard