Merge Two Sorted Lists

To merge two sorted linked lists, you can use a recursive approach or an iterative approach with a dummy node. Here’s an example of the iterative approach, which is often easier to understand for beginners:

Initialize a dummy node to act as the starting point of the merged list.
Use two pointers, one for each of the given lists, to iterate through the nodes.
Compare the values of the nodes the pointers are pointing to, and link the smaller one to the merged list.
Move the pointer of the linked node to the next node in its list.
Repeat steps 3-4 until one of the lists is fully traversed.
If there are any remaining nodes in the other list, link them to the merged list.
Return the node next to the dummy node, which is the head of the merged list.

Here’s the code:

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class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        # Initialize a dummy node
        dummy = ListNode(-1)
        prev = dummy

        # Iterate through both lists until one of them is empty
        while list1 and list2:
            if list1.val < list2.val:
                prev.next = list1
                list1 = list1.next
            else:
                prev.next = list2
                list2 = list2.next

            # Move to the next node in the merged list
            prev = prev.next

        # If there are any remaining nodes in either list, link them to the merged list
        prev.next = list1 if list1 else list2

        # Return the head of the merged list (the node next to the dummy node)
        return dummy.next

The time complexity of this solution is (O(n + m)), where (n) and (m) are the lengths of the two lists. The space complexity is (O(1)), as we only use a constant amount of extra space.

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class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode()
        while list1 and list2:               
            if list1.val < list2.val:
                cur.next = list1
                list1, cur = list1.next, list1
            else:
                cur.next = list2
                list2, cur = list2.next, list2

        if list1 or list2:
            cur.next = list1 if list1 else list2

        return dummy.next

Problem Classification

Data Structure: The problem is heavily related to the Linked List data structure, as it involves operations like merging, and traversing through nodes.
Algorithm: The problem also involves algorithmic components related to pointer manipulation and traversal.
Domain: The problem falls into the domain of list manipulation, as it requires dealing with data nodes and links between them.

The goal is to merge two sorted linked lists into one sorted linked list. This requires a good understanding of Linked Lists and iterative data processing.

Language Agnostic Coding Drills

The problem is about merging two sorted linked lists. The code provided uses a dummy node to anchor the resulting linked list and a cursor node that always points to the last node in the result.

Understanding of linked lists: Before diving into the problem, it’s crucial to have a solid understanding of linked lists - a data structure where elements are connected through references (or ’links’). Learn how to create a linked list, how to traverse a linked list, and how to manipulate the ’next’ pointers.
Understanding of merge algorithm: Understand how to merge two sorted arrays/lists. You would typically have two pointers, each pointing to the start of the two lists. At each step, you compare the elements each pointer is pointing to, and pick the smaller one to add to the result.
Dummy node usage: Understand the usage of a dummy node in linked list manipulation. A dummy node is a technique used to simplify edge cases where the head of a linked list might change.
Iterative traversal of linked lists: Familiarize yourself with iterative methods to traverse through a linked list using while loops.
Pointer manipulation: The crux of this problem lies in correctly manipulating the ’next’ pointers of the lists and the nodes in the new list. It requires a solid understanding of how pointers/references work in your chosen programming language.

In terms of problem-solving approach, the steps would be:

Start by initializing a dummy ListNode and a cursor that points to this dummy node.
Enter a loop that continues as long as there are nodes in both lists.
In each iteration of the loop, compare the current nodes in each list, append the node with the smaller value to the result list, and move the pointer of the list with the smaller element forward.
If one list becomes empty before the other, append the rest of the non-empty list to the result list.
Finally, return the next node of the dummy node (as the dummy was just a placeholder).

Targeted Drills in Python

Drill 1 - Understanding Linked Lists

A linked list is a sequence of data structures, which are connected together via links. Here is a simple implementation of a Node and creating a linked list using these Nodes.

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

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

    def add(self, data):
        if not self.head:
            self.head = Node(data)
        else:
            cur = self.head
            while cur.next:
                cur = cur.next
            cur.next = Node(data)

# Create a linked list and add data
ll = LinkedList()
ll.add(1)
ll.add(2)
ll.add(3)

Drill 2 - Understanding Merge Algorithm

This drill focuses on the merge operation, assuming that we have two sorted lists.

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def merge_lists(list1, list2):
    result = []
    i = j = 0
    while i < len(list1) and j < len(list2):
        if list1[i] < list2[j]:
            result.append(list1[i])
            i += 1
        else:
            result.append(list2[j])
            j += 1
    while i < len(list1):
        result.append(list1[i])
        i += 1
    while j < len(list2):
        result.append(list2[j])
        j += 1
    return result

# Test the function
print(merge_lists([1,3,5], [2,4,6]))  # Output: [1, 2, 3, 4, 5, 6]

Drill 3 - Understanding Dummy Node Usage

The usage of a dummy node can simplify the process of manipulating the head node. This drill will demonstrate the usage of a dummy node.

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

def add_node(head, data):
    dummy = Node()  
    dummy.next = head
    cur = dummy  
    while cur.next:
        cur = cur.next
    cur.next = Node(data)  
    return dummy.next

# Test the function
head = Node(1)
head.next = Node(2)
head = add_node(head, 3)
print(head.data, head.next.data, head.next.next.data)  # Output: 1 2 3

Drill 4 - Iterative traversal of linked lists

This drill will focus on traversing a linked list in an iterative manner.

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

def print_list(head):
    cur = head
    while cur:
        print(cur.data, end=' ')
        cur = cur.next

# Test the function
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
print_list(head)  # Output: 1 2 3

Drill 5 - Pointer Manipulation

This final drill focuses on pointer manipulation, which is key to solving this problem.

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

def swap_adjacent_nodes(head):
    dummy = Node(0)
    dummy.next = head
    prev_node = dummy

    while head and head.next:
        first_node = head
        second_node = head.next
        prev