<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>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

def calc_divisor(a, b):
    maxv = max(a,b)
    minv = min(a,b)

    while maxv % minv != 0:
        modv = maxv % minv
        maxv = max(modv, minv)
        minv = min(modv, minv)

    return minv


    # if mod == 0:
    #     return minv
    # else:
    #     return calc_divisor(mod, minv)

class Solution:
    def insertGreatestCommonDivisors(self, head: Optional[ListNode]) -> Optional[ListNode]:
        
        tail = head
        lead = tail.next

        if not lead:
            return tail

        lval = lead.val
        tval = tail.val

        # divisor = 1
        # minval = min(lval, tval)
        # for i in range(minval):
        #     # print(lval, tval)
        #     # print(i)
        #     if lval % (minval-i) == 0 and tval % (minval-i) == 0:
        #         divisor = minval-i
        #         break
        #         # print('divisor',divisor)
        # divisor = calc_divisor(lval, tval)

        maxv = max(lval,tval)
        minv = min(lval,tval)

        while maxv % minv != 0:
            modv = maxv % minv
            maxv = max(modv, minv)
            minv = min(modv, minv)

        node = ListNode(val=minv)
        node.next = lead
        tail.next = node

        while lead.next:
            tail = lead
            lead = lead.next

            lval = lead.val
            tval = tail.val

            # divisor = 1
            # minval = min(lval, tval)
            # for i in range(minval):
            #     if lval % (minval-i) == 0 and tval % (minval-i) == 0:
            #         divisor = minval-i
            #         break
            # divisor = calc_divisor(lval, tval)
            maxv = max(lval,tval)
            minv = min(lval,tval)

            while maxv % minv != 0:
                modv = maxv % minv
                maxv = max(modv, minv)
                minv = min(modv, minv)


            node = ListNode(val=minv)
            node.next = lead
            tail.next = node

        return head
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