Lesson 48: Mastering JavaScript Linked list with challenges!

A linked list is a linear data structure made of nodes, where each node has:
value
: the data the node holdsnext
: a reference to the next node (ornull
if it's the last node)
let node = {
value: 10,
next: null
};
The head node is the entry point. Traversing requires walking node-by-node.
🆚 Compared to Arrays
Feature | Array | Linked List |
Indexed Access | Fast (O(1) ) | Slow (O(n) ) |
Insert at Beginning | Slow (O(n) ) | Fast (O(1) ) |
Delete at Beginning | Slow (O(n) ) | Fast (O(1) ) |
Memory locality | Contiguous | Non-contiguous |
Enhancements
Doubly Linked List: adds
prev
pointerTail Pointer: fast access to last node
Circular Linked List: last node points back to head
📉 2. Fill Gaps — Hidden Corners & Traps
list.next
=
list.next.next
skips over a node. If you're holding no references to the skipped node, it's eligible for garbage collection.Be cautious with circular references — they can cause infinite loops if not handled (e.g.,
node.next
= node
).Traversing a list is inherently recursive in structure — can be implemented iteratively or recursively, but deep recursion can cause stack overflow.
💻 3. Challenge Me Deeply — 10 Expert-Level Practice Problems
🧩 Challenge 1: Reverse a Singly Linked List (Iteratively)
function reverseList(head) {
let prev = null, current = head;
while (current) {
const next = current.next;
current.next = prev;
prev = current;
current = next;
}
return prev;
}
🧩 Challenge 2: Detect a Cycle (Floyd's Tortoise and Hare)
function hasCycle(head) {
let slow = head, fast = head;
while (fast && fast.next) {
slow = slow.next;
fast = fast.next.next;
if (slow === fast) return true;
}
return false;
}
🧩 Challenge 3: Find Middle Node
function middleNode(head) {
let slow = head, fast = head;
while (fast && fast.next) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
🧩 Challenge 4: Merge Two Sorted Linked Lists
function mergeLists(l1, l2) {
if (!l1) return l2;
if (!l2) return l1;
if (l1.value < l2.value) {
l1.next = mergeLists(l1.next, l2);
return l1;
} else {
l2.next = mergeLists(l1, l2.next);
return l2;
}
}
🧩 Challenge 5: Remove N-th Node From End
function removeNthFromEnd(head, n) {
let dummy = { next: head };
let fast = dummy, slow = dummy;
for (let i = 0; i <= n; i++) fast = fast.next;
while (fast) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return dummy.next;
}
🧩 Challenge 6: Check if a List is a Palindrome
function isPalindrome(head) {
let slow = head, fast = head;
const stack = [];
while (fast && fast.next) {
stack.push(slow.value);
slow = slow.next;
fast = fast.next.next;
}
if (fast) slow = slow.next;
while (slow) {
if (stack.pop() !== slow.value) return false;
slow = slow.next;
}
return true;
}
🧩 Challenge 7: Clone a Linked List With Random Pointers (advanced)
function copyRandomList(head) {
let map = new Map();
let current = head;
while (current) {
map.set(current, { value: current.value });
current = current.next;
}
current = head;
while (current) {
let clone = map.get(current);
clone.next = map.get(current.next) || null;
clone.random = map.get(current.random) || null;
current = current.next;
}
return map.get(head);
}
🧩 Challenge 8: Convert Array to Linked List and vice versa
function arrayToList(arr) {
let head = null;
for (let i = arr.length - 1; i >= 0; i--) {
head = { value: arr[i], next: head };
}
return head;
}
function listToArray(head) {
const result = [];
while (head) {
result.push(head.value);
head = head.next;
}
return result;
}
🧩 Challenge 9: Recursive Length of List
function getLength(node) {
return node ? 1 + getLength(node.next) : 0;
}
🧩 Challenge 10: Find Intersection of Two Lists
function getIntersection(l1, l2) {
let a = l1, b = l2;
while (a !== b) {
a = a ? a.next : l2;
b = b ? b.next : l1;
}
return a; // or null
}
💼 4. Interview-Ready Questions
Conceptual
Q: Why are linked lists better for queues than arrays?
A: Linked lists allowO(1)
insertion/removal at both ends without reindexing.Q: When would you prefer a doubly linked list over singly?
A: When you need to traverse both forward and backward efficiently.Q: Why can’t we access the N-th element instantly in a linked list?
A: Because each node only knows about the next — you must traverse linearly.
Coding
Implement a linked list class with
append
,prepend
,delete
,find
, andtoArray
methods.Detect and remove a cycle in a linked list.
Design a real-time chat buffer using linked list to avoid expensive array operations.
🌍 5. Real-World Usage Examples
Undo/Redo Stack in editors: typically implemented via doubly linked list.
Queue systems (e.g., job schedulers, print queues).
Browser history navigation.
Streaming data (audio/video): when buffering using constant-time appends/removals.
🧠 6. Memory Aids & Visualization
Linked List Mnemonic
"Next Stops Point the Way"
🧠 Think: Every node is a bus stop, andnext
points to the next destination.
ASCII Diagram
[1] → [2] → [3] → [4] → null
🎮 7. Fun Project: Interactive Linked List Playground
Build a web-based Linked List Visualizer in JavaScript:
Add, remove, reverse nodes
Animate
next
pointersShow traversal and mid-search
Toggle singly/doubly/circular mode
Frameworks: HTML/CSS/Canvas or React + D3.js
✅ Pro Tips
Avoid linked lists for random-access tasks — arrays or typed arrays are better.
For extremely large data pipelines (streaming, processing) where append/remove dominate, linked lists save memory churn.
Use
null
carefully — a missed base case in recursive linked list problems can cause stack overflow or infinite loops.
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