Reverse Nodes in k-Group

You are asked to reverse the nodes of a linked list in groups of size ( k ). If the number of nodes is not a multiple of ( k ), the remaining nodes should not be reversed.

Here’s a simple approach:

  1. Calculate the Length: First, you’ll need to calculate the length of the linked list to know how many groups of ( k ) there are.
  2. Reverse k Groups: Iterate through the list and reverse every ( k ) nodes. You need to handle the connections carefully while reversing.
  3. Connect Remaining Nodes: If the total number of nodes is not a multiple of ( k ), you’ll need to connect the remaining nodes without reversing.

Here’s the code implementing the above steps:

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class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        # Calculate the length of the linked list
        length = 0
        temp = head
        while temp:
            length += 1
            temp = temp.next

        # Dummy node to simplify the code
        dummy = ListNode(0)
        dummy.next = head
        prev_group_end = dummy

        while length >= k:
            # Reversing k nodes and handling the connections
            current_group_start = prev_group_end.next
            current_group_end = prev_group_end.next
            for _ in range(k - 1):
                current_group_end = current_group_end.next
            next_group_start = current_group_end.next

            prev = None
            curr = current_group_start
            for _ in range(k):
                next_node = curr.next
                curr.next = prev
                prev = curr
                curr = next_node

            # Connecting the reversed group with the previous group
            prev_group_end.next = current_group_end
            current_group_start.next = next_group_start
            prev_group_end = current_group_start

            # Reduce the length
            length -= k

        return dummy.next

Explanation:

  • You iterate through the linked list, reversing groups of ( k ) nodes and handling the connections between groups.
  • Dummy node helps in simplifying the code by providing a previous node to the head, which makes the connection easier.

This code has a time complexity of ( O(n) ), where ( n ) is the number of nodes in the linked list, and it meets the constraints of the problem.

10 Prerequisite LeetCode Problems

The problem “25. Reverse Nodes in k-Group” involves advanced linked list manipulation. Here are 10 problems to prepare you:

  1. “Delete Node in a Linked List” (LeetCode Problem #237): This is a basic problem to get you familiar with how to manipulate nodes in a linked list.

  2. “Reverse Linked List” (LeetCode Problem #206): This problem is the basis for the Reverse Nodes in k-Group problem. Understanding this is crucial.

  3. “Middle of the Linked List” (LeetCode Problem #876): This problem helps improve your skill in traversing linked lists.

  4. “Palindrome Linked List” (LeetCode Problem #234): This problem can help you understand how to traverse and manipulate a linked list in more complex ways.

  5. “Remove Linked List Elements” (LeetCode Problem #203): This problem will help you understand how to remove nodes from a linked list based on certain conditions.

  6. “Remove Nth Node From End of List” (LeetCode Problem #19): This problem is a slightly trickier linked list manipulation problem that will help you handle edge cases in linked list manipulation.

  7. “Swap Nodes in Pairs” (LeetCode Problem #24): This is a precursor to the k-Group problem, where you swap nodes in pairs. This will help you understand the mechanics of how you’ll need to reverse nodes in groups of k.

  8. “Odd Even Linked List” (LeetCode Problem #328): This problem can improve your ability to manipulate pointers in a linked list in a more complex way.

  9. “Partition List” (LeetCode Problem #86): This problem requires you to partition a linked list around a value, which can be good practice for pointer manipulation.

  10. “Add Two Numbers” (LeetCode Problem #2): This problem is a more complex linked list manipulation problem that will test your ability to handle edge cases and manipulate pointers.

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        curr = head
        for _ in range(k):
            if not curr: return head
            curr = curr.next
		        
        prev = None
        curr = head
        for _ in range(k):
            nxt = curr.next
            curr.next = prev
            prev = curr
            curr = nxt
        		
        head.next = self.reverseKGroup(curr, k)
        return prev

Problem Classification

This problem can be classified as a Grouping and Rearrangement problem. Here, we are given a sequence (which could represent a lineup, queue, etc.) and we’re asked to group the items into subsets of a specific size, and then rearrange (reverse) the order of items within those groups. This could potentially be applied to problems in logistics or scheduling, where certain tasks or items need to be grouped together and processed in a specific order.

Language Agnostic Coding Drills

  1. Understanding Variables and Basic Data Structures: This concept is all about knowing how to declare and use basic data structures like integers, strings, and especially here, classes to create a singly-linked list.

  2. Object-Oriented Programming: This concept involves understanding how to create classes, and in this context, create a class Solution that contains a method reverseKGroup. This also includes understanding how to create an instance method and understanding self.

  3. Understanding linked list: Understanding the structure and properties of a linked list is critical here. A linked list is a linear collection of data elements, each pointing to the next, which forms a sequence or chain.

  4. Understanding linked list reversal: Understanding the concept of how to reverse a linked list. This includes knowing how to keep track of the current, previous, and next nodes.

  5. Recursion: Understanding how to use recursion. Recursion is when a function calls itself in its definition. Here, reverseKGroup is a recursive function that calls itself.

  6. Looping through a linked list: This is about knowing how to traverse a linked list, which can be done using a for loop.

Now, here’s the problem-solving approach step by step:

The main goal of the reverseKGroup function is to reverse nodes in a linked list in k-group. If the number of nodes is not a multiple of k, then the remaining nodes are left as it is.

  1. Firstly, check if there are k nodes left from the current head node. If there are not enough nodes left, simply return the head without doing anything.

  2. If there are at least k nodes, then we reverse these k nodes. This process involves keeping track of the current, previous, and next nodes. After reversing k nodes, head now becomes the last node in this group, and curr points to the first node of the next group (or null).

  3. The reversed group also needs to be connected back to the rest of the list. For that, we recursively call the reverseKGroup function for the remaining list (i.e., curr), and connect the original head node to the result.

  4. Finally, return prev, which is the new head of this group after the reversal.

Therefore, the final solution is derived by recursively applying the reversal of the linked list on k-group and then linking them back to form a complete list.

Targeted Drills in Python

Make sure you understand and can do these exercises, as they are key building blocks of the final solution:

  1. Understanding Variables and Basic Data Structures:

    Understanding of basic variables and data structures can be tested with a simple task, such as creating a class to represent a singly-linked list.

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    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

  2. Object-Oriented Programming:

    Understanding of object-oriented programming can be drilled with creating classes and methods.

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    class Solution:
    def helloWorld(self):
        print("Hello, World!")

  3. Understanding linked list:

    Traverse through a linked list and print each node’s value.

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    def print_list(node):
    while node:
        print(node.val)
        node = node.next

  4. Understanding linked list reversal:

    Write a function to reverse a linked list.

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    def reverse_list(node):
    prev = None
    while node:
        next_node = node.next
        node.next = prev
        prev = node
        node = next_node
    return prev

  5. Recursion:

    Using recursion to solve a problem, such as finding the factorial of a number.

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    def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

  6. Looping through a linked list:

    Using a loop to traverse through a linked list, such as finding the length of the linked list.

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    def find_length(node):
    count = 0
    while node:
        count += 1
        node = node.next
    return count

After you’re comfortable with these concepts, you should be able to better understand and implement the final solution. The final solution uses these concepts together - it uses a loop to traverse through the linked list and reverses groups of nodes recursively.