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    Radix Sort

Linked List Visualization

A Linked List is a linear data structure where elements are stored in nodes, and each node points to the next node in the sequence. Unlike arrays, linked lists do not have a fixed size and can grow or shrink dynamically. This makes them efficient for insertions and deletions, but slower for direct access to an element.

Visualizer



Pseudocode

function insertAtHead(value):
    node = new Node(value)
    node.next = head.next
    head.next = node
function insertAtTail(value):
    node = new Node(value)
    cur = head
    while cur.next is not null:
        cur = cur.next
    cur.next = node
function insertAt(index, value):
    if index == 0:
        insertAtHead(value)
        return
    cur = head
    for i = 1 to index:
        if cur.next is null: break
        cur = cur.next
    node = new Node(value)
    node.next = cur.next
    cur.next = node
function deleteAt(index):
    cur = head
    prev = null
    for i = 0 to index:
        prev = cur
        cur = cur.next
    prev.next = cur.next



Common Interview Questions

Compare array access vs Linked List access in terms of memory cache performance.

Arrays store elements in contiguous memory blocks, enabling O(1) index access and excellent CPU spatial cache prefetching. Linked list nodes are allocated dynamically across heap memory, leading to memory fragmentation and frequent CPU cache misses.

How do you reverse a Linked List in-place in O(n) time and O(1) space?

Maintain three pointers: prev (initialized to null), curr (head), and next. Iterate through the list:

Code Snippet
let prev = null, cur = head;
while (cur) {
  let next = cur.next;
  cur.next = prev;
  prev = cur;
  cur = next;
}
head = prev;
How do you detect a cycle in a Linked List using Floyd's Cycle-Finding algorithm (Tortoise and Hare)?

Use two pointers: slow moving 1 step at a time, and fast moving 2 steps at a time. If the list has a cycle, fast will eventually catch up and meet slow inside the loop. If fast reaches null, no cycle exists.

Code Snippet
let slow = head, fast = head;
while (fast && fast.next) {
  slow = slow.next;
  fast = fast.next.next;
  if (slow == fast) return true; // Cycle detected
}
return false;

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