<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
class Solution:
    def insertGreatestCommonDivisors(self, head: Optional[ListNode]) -> Optional[ListNode]:
        self.lookup = defaultdict()

        def findFactors(num):
            factors = []
            for f in range(2, num+1):
                while num%f == 0:
                    factors.append(f)
                    num = num//f
            
            return factors

        def findGCD(f1, f2):
            commonFactors = 1
            ptr1, ptr2 = 0,0

            while ptr1 < len(f1) and ptr2 < len(f2):
                if f1[ptr1] == f2[ptr2]:
                    commonFactors*=f1[ptr1]
                    ptr1+=1
                    ptr2+=1

                else:
                    if f1[ptr1] < f2[ptr2]:
                        ptr1+=1

                    else:
                        ptr2+=1

            return commonFactors

        curr = head
        while curr:
            if curr.val not in self.lookup:
                self.lookup[curr.val] = findFactors(curr.val)
            curr = curr.next

        # print(self.lookup)

        prev = head
        curr = head.next

        while curr:
            gcd = findGCD(self.lookup[prev.val], self.lookup[curr.val])
            new_node = ListNode(gcd)
            prev.next = new_node
            new_node.next = curr
            prev = curr
            curr = curr.next

        return head




        
                
                     
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