LINKED LISTS
============
A linked list stores elements in nodes. Each node holds data and a pointer to the next node.
Unlike arrays, nodes are not stored in contiguous memory.

Advantages: O(1) insert/delete at head, dynamic size.
Disadvantages: O(n) access by index, extra memory for pointers.

--- Node Definition ---
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

--- Singly Linked List ---
Each node points to the next. Last node points to None.

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_at_head(self, data):
        node = Node(data)
        node.next = self.head
        self.head = node

    def insert_at_tail(self, data):
        node = Node(data)
        if not self.head:
            self.head = node
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = node

    def delete(self, data):
        if not self.head:
            return
        if self.head.data == data:
            self.head = self.head.next
            return
        curr = self.head
        while curr.next and curr.next.data != data:
            curr = curr.next
        if curr.next:
            curr.next = curr.next.next

    def search(self, data):
        curr = self.head
        while curr:
            if curr.data == data:
                return True
            curr = curr.next
        return False

    def print_list(self):
        curr = self.head
        result = []
        while curr:
            result.append(curr.data)
            curr = curr.next
        print(result)

ll = LinkedList()
ll.insert_at_tail(1)
ll.insert_at_tail(2)
ll.insert_at_tail(3)
ll.insert_at_head(0)
ll.print_list()   # [0, 1, 2, 3]
ll.delete(2)
ll.print_list()   # [0, 1, 3]

--- Reverse a Linked List ---
Time: O(n). Space: O(1).

def reverse(head):
    prev = None
    curr = head
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    return prev  # new head

--- Detect Cycle (Floyd's Algorithm) ---
Use slow and fast pointers. If they meet, there is a cycle.
Time: O(n). Space: O(1).

def has_cycle(head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow == fast:
            return True
    return False

--- Find Middle of Linked List ---
Time: O(n). Space: O(1).

def find_middle(head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    return slow  # slow is at middle

--- Merge Two Sorted Linked Lists ---
Time: O(n + m). Space: O(1).

def merge_sorted(l1, l2):
    dummy = Node(0)
    curr = dummy
    while l1 and l2:
        if l1.data <= l2.data:
            curr.next = l1
            l1 = l1.next
        else:
            curr.next = l2
            l2 = l2.next
        curr = curr.next
    curr.next = l1 or l2
    return dummy.next

--- Remove Nth Node From End ---
Use two pointers separated by n steps.
Time: O(n). Space: O(1).

def remove_nth_from_end(head, n):
    dummy = Node(0)
    dummy.next = head
    fast = slow = dummy
    for _ in range(n + 1):
        fast = fast.next
    while fast:
        slow = slow.next
        fast = fast.next
    slow.next = slow.next.next
    return dummy.next

--- Check if Palindrome ---
Time: O(n). Space: O(1).

def is_palindrome(head):
    # Find middle
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    # Reverse second half
    prev = None
    curr = slow
    while curr:
        nxt = curr.next
        curr.next = prev
        prev = curr
        curr = nxt
    # Compare halves
    left, right = head, prev
    while right:
        if left.data != right.data:
            return False
        left = left.next
        right = right.next
    return True

--- Doubly Linked List ---
Each node has a pointer to both next and previous nodes.

class DNode:
    def __init__(self, data):
        self.data = data
        self.next = None
        self.prev = None

class DoublyLinkedList:
    def __init__(self):
        self.head = None

    def insert_at_head(self, data):
        node = DNode(data)
        if self.head:
            self.head.prev = node
        node.next = self.head
        self.head = node

    def delete(self, node):
        if node.prev:
            node.prev.next = node.next
        else:
            self.head = node.next
        if node.next:
            node.next.prev = node.prev

--- Circular Linked List ---
Last node points back to head forming a circle.

def is_circular(head):
    if not head:
        return False
    curr = head.next
    while curr and curr != head:
        curr = curr.next
    return curr == head