<p>Given the head of a linked list <code>head</code>, in which each node contains an integer value.</p>

<p>Between every pair of adjacent nodes, insert a new node with a value equal to the <strong>greatest common divisor</strong> of them.</p>

<p>Return <em>the linked list after insertion</em>.</p>

<p>The <strong>greatest common divisor</strong> of two numbers is the largest positive integer that evenly divides both numbers.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/18/ex1_copy.png" style="width: 641px; height: 181px;" />
<pre>
<strong>Input:</strong> head = [18,6,10,3]
<strong>Output:</strong> [18,6,6,2,10,1,3]
<strong>Explanation:</strong> The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).
- We insert the greatest common divisor of 18 and 6 = 6 between the 1<sup>st</sup> and the 2<sup>nd</sup> nodes.
- We insert the greatest common divisor of 6 and 10 = 2 between the 2<sup>nd</sup> and the 3<sup>rd</sup> nodes.
- We insert the greatest common divisor of 10 and 3 = 1 between the 3<sup>rd</sup> and the 4<sup>th</sup> nodes.
There are no more adjacent nodes, so we return the linked list.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/18/ex2_copy1.png" style="width: 51px; height: 191px;" />
<pre>
<strong>Input:</strong> head = [7]
<strong>Output:</strong> [7]
<strong>Explanation:</strong> The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes.
There are no pairs of adjacent nodes, so we return the initial linked list.
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 5000]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 1000</code></li>
</ul>

from typing import Optional


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


class Solution:

    def gcd(a, b):

        # GCD(0, b) == b; GCD(a, 0) == a,
        # GCD(0, 0) == 0
        if a == 0:
            return b

        if b == 0:
            return a

        # Finding K, where K is the
        # greatest power of 2 that
        # divides both a and b.
        k = 0

        while ((a | b) & 1) == 0:
            a = a >> 1
            b = b >> 1
            k = k + 1

        # Dividing a by 2 until a becomes odd
        while (a & 1) == 0:
            a = a >> 1

        # From here on, 'a' is always odd.
        while b != 0:

            # If b is even, remove all
            # factor of 2 in b
            while (b & 1) == 0:
                b = b >> 1

            # Now a and b are both odd. Swap if
            # necessary so a <= b, then set
            # b = b - a (which is even).
            if a > b:

                # Swap u and v.
                temp = a
                a = b
                b = temp

            b = b - a

        # restore common factors of 2
        return a << k

    def insertGreatestCommonDivisors(
        self, head: Optional[ListNode]
    ) -> Optional[ListNode]:
        if not head or not head.next:
            return head

        current = head
        while current.next:
            gcd_value = gcd(current.val, current.next.val)
            gcd_node = ListNode(gcd_value)
            gcd_node.next = current.next
            current.next = gcd_node
            current = gcd_node.next

        return head

19200