<p>Given the <code>head</code> of a singly linked list, return <code>true</code><em> if it is a </em><span data-keyword="palindrome-sequence"><em>palindrome</em></span><em> or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal1linked-list.jpg" style="width: 422px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2,2,1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal2linked-list.jpg" style="width: 182px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do it in <code>O(n)</code> time and <code>O(1)</code> space?
'''
reevrse seconf half 
iterat first and sencond half togater coparing vals

time = o(n)
space = o(1)
'''

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return True

        slow = head
        fast = head.next
        # first_half = slow # used in last step
        # first_half will be long in cse of odd
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        
        second_half = slow.next
        
        prev = None
        while second_half:
            temp = second_half.next
            second_half.next = prev
            prev = second_half
            second_half = temp
        print(prev.val)
        
        slow.next = None
        second_half = prev
        first_half = head

        while second_half:
            print(second_half.val, first_half.val)
            if second_half.val != first_half.val:
                return False
            second_half = second_half.next  
            first_half = first_half.next
        
        count = 0
        while first_half:
            first_half = first_half.next

        return True if count <= 1 else False
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