Introduction

In this blog post, we will explore how to solve the LeetCode problem "25" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is. You may not alter the values in the list's nodes, only nodes themselves may be changed. Example 1: Input: head = [1,2,3,4,5], k = 2 Output: [2,1,4,3,5] Example 2: Input: head = [1,2,3,4,5], k = 3 Output: [3,2,1,4,5] Constraints: The number of nodes in the list is n. 1 <= k <= n <= 5000 0 <= Node.val <= 1000 Follow-up: Can you solve the problem in O(1) extra memory space?

Explanation

Here's a solution to the k-group reverse linked list problem, focusing on efficiency and clarity.

• Iterative Reversal: The solution uses an iterative approach to reverse groups of k nodes. It maintains pointers to the previous group's tail, the current group's head, and performs local reversals within each group.
• Connectivity: After reversing each group, the code carefully connects the reversed group to the previous and next parts of the list.

• Handling Insufficient Nodes: If a group has fewer than k nodes, it's left untouched.

• Time & Space Complexity: O(n) time complexity, O(1) space complexity.

Code

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

def reverseKGroup(head, k):
    """
    Reverses the nodes of a linked list k at a time.

    Args:
        head: The head of the linked list.
        k: The size of the group to reverse.

    Returns:
        The head of the modified linked list.
    """
    dummy = ListNode(0)  # Dummy node to handle the head case
    dummy.next = head
    prev_group_tail = dummy

    while True:
        # Check if there are at least k nodes remaining
        current_group_head = prev_group_tail.next

        #Temporary pointer to iterate over k nodes, if k nodes not present, break
        temp = current_group_head
        for _ in range(k):
            if not temp:
                return dummy.next  # Return if fewer than k nodes remain
            temp = temp.next

        # Reverse the k nodes
        prev = None
        current = current_group_head
        for _ in range(k):
            next_node = current.next
            current.next = prev
            prev = current
            current = next_node

        # Connect the reversed group to the previous and next groups
        prev_group_tail.next = prev  # prev is now the head of the reversed group
        current_group_head.next = current  # current is now the node after the reversed group

        #Update the prev group tail to point to last node of last reversed group
        prev_group_tail = current_group_head # update prev_group_tail for the next group