Understanding the Problem

The task involves modifying a linked list by reversing its nodes in groups of k. Given the head of a linked list, the goal is to reverse the nodes k at a time and return the updated list. If the total number of nodes is not a multiple of k, the remaining nodes at the end should stay in their original order. The values within the nodes cannot be altered—only the nodes themselves can be rearranged.

Example Scenarios:

1. Input: head = [1,2,3,4,5], k = 2

   Output: [2,1,4,3,5]

2. Input: head = [1,2,3,4,5], k = 3

   Output: [3,2,1,4,5]

Constraints:

• The number of nodes, n, must satisfy 1 <= k <= n <= 50000.
• Node values must be in the range 0 <= Node.val <= 1000.

Follow-Up: Can this problem be solved using O(1) extra memory?

Intuition

Utilizing recursion fits naturally with this problem since it allows operations on fixed segments of the linked list incrementally. Each segment of k nodes can be treated as an independent linked list. The main focus is reversing the k nodes while ensuring that connections are properly established as the recursion unwinds.

Algorithm Steps

1. Define a reverse() function that accepts the head of the linked list and an integer k.
2. Count the nodes until k is reached.
3. If fewer than k nodes remain, return the current head.
4. If at least k nodes are present, reverse the first k nodes and recursively call the function on the remaining list.

Here’s a brief dry run of the algorithm. For a linked list such as 1,2,3,4,5 with k=2:

• Reverse the first two nodes to get 2,1.
• The recursion will then handle 3,4,5, producing a final output of 2,1,4,3,5.

class Solution:

def reverseLinkedList(self, head, k):

new_head, ptr = None, head

while k:

next_node = ptr.next

ptr.next = new_head

new_head = ptr

ptr = next_node

k -= 1

return new_head

def reverseKGroup(self, head: ListNode, k: int) -> ListNode:

count = 0

ptr = head

while count < k and ptr:

ptr = ptr.next

count += 1

if count == k:

reversedHead = self.reverseLinkedList(head, k)

head.next = self.reverseKGroup(ptr, k)

return reversedHead

return head

Visual Learning

To enhance understanding, here are two informative videos:

1. Reverse Nodes in K-Group - Linked List - Leetcode 25

   This video provides a comprehensive walkthrough of the algorithm with visual aids to clarify the reversal process.

2. How to Reverse a LinkedList In-place by Alternating K Sublists

   This video delves into an in-place technique for reversing linked lists, offering an alternative perspective on the problem.

Intuition

The iterative method mirrors the recursive approach but eliminates the need for a call stack by utilizing a few extra pointers to maintain connections. This method also processes k nodes at a time and ensures connections are established as nodes are reversed.

Algorithm Steps

1. Maintain pointers for the head, new head (tail of the reversed segment), tail (previous segment), and the final output head.
2. Count k nodes and reverse them as before.
3. If the current k segment is reversed, link it to the previous segment.
4. Repeat until all nodes are processed.

Here’s a concise implementation of the iterative solution:

class Solution:

def reverseLinkedList(self, head, k):

new_head, ptr = None, head

while k:

next_node = ptr.next

ptr.next = new_head

new_head = ptr

ptr = next_node

k -= 1

return new_head

def reverseKGroup(self, head: ListNode, k: int) -> ListNode:

ptr = head

ktail = None

new_head = None

while ptr:

count = 0

temp_head = head

while count < k and ptr:

ptr = ptr.next

count += 1

if count == k:

revHead = self.reverseLinkedList(head, k)

if not new_head:

new_head = revHead

if ktail:

ktail.next = revHead

ktail = temp_head

head = ptr

if ktail:

ktail.next = head

return new_head if new_head else head

Conclusion

Both approaches yield an efficient solution to the problem, with time complexity O(N) and space complexity O(N) for the recursive method, and O(1) for the iterative method.

Stay tuned for more intriguing interview questions and strategies as we continue to explore the world of algorithms!

— I am a senior software engineer at MANNG. Follow me for more engaging content on tech interviews and algorithm strategies.