# Solving Leetcode Interviews in Seconds with AI: Odd Even Linked List


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "328" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list. The first node is considered odd, and the second node is even, and so on. Note that the relative order inside both the even and odd groups should remain as it was in the input. You must solve the problem in O(1) extra space complexity and O(n) time complexity.   Example 1:   Input: head = [1,2,3,4,5] Output: [1,3,5,2,4]  Example 2:   Input: head = [2,1,3,5,6,4,7] Output: [2,3,6,7,1,5,4]    Constraints:  The number of nodes in the linked list is in the range [0, 104]. -106 <= Node.val <= 106  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **Approach:**
    *   Maintain two pointers, one for odd nodes and one for even nodes.
    *   Iterate through the list, connecting odd nodes together and even nodes together.
    *   Finally, connect the tail of the odd nodes to the head of the even nodes.

*   **Complexity:**
    *   O(n) time complexity, O(1) space complexity

	
	# Code
	```python
	class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def oddEvenList(head: ListNode) -> ListNode:
    if not head:
        return head

    odd = head
    even = head.next
    even_head = even

    while even and even.next:
        odd.next = even.next
        odd = odd.next
        even.next = odd.next
        even = even.next

    odd.next = even_head

    return head
	```