<p>You are given two integers <code>m</code> and <code>n</code>, which represent the dimensions of a matrix.</p>

<p>You are also given the <code>head</code> of a linked list of integers.</p>

<p>Generate an <code>m x n</code> matrix that contains the integers in the linked list presented in <strong>spiral</strong> order <strong>(clockwise)</strong>, starting from the <strong>top-left</strong> of the matrix. If there are remaining empty spaces, fill them with <code>-1</code>.</p>

<p>Return <em>the generated matrix</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/09/ex1new.jpg" style="width: 240px; height: 150px;" />
<pre>
<strong>Input:</strong> m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
<strong>Output:</strong> [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
<strong>Explanation:</strong> The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/11/ex2.jpg" style="width: 221px; height: 60px;" />
<pre>
<strong>Input:</strong> m = 1, n = 4, head = [0,1,2]
<strong>Output:</strong> [[0,1,2,-1]]
<strong>Explanation:</strong> The diagram above shows how the values are printed from left to right in the matrix.
The last space in the matrix is set to -1.</pre>

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

<ul>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li>The number of nodes in the list is in the range <code>[1, m * n]</code>.</li>
	<li><code>0 &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 spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]:

        pos = [0,0] #[y, x], top left is [0,0]
        results = [[-1 for j in range(n)] for i in range(m)]
        current = head
        counter = 0
        total = m*n

        def move_north():
            pos[0] -= 1
        def move_east():
            pos[1] += 1
        def move_south():
            pos[0] += 1
        def move_west():
            pos[1] -= 1
        def update_results():
            nonlocal results, current
            if current:
                results[pos[0]][pos[1]] = current.val
                current = current.next
                return True
            else:
                return False

        if m * n == 1:  # easy edge case to save resources
            return [[head.val]]        
        
        while current and counter <= total:
            traversed = False
            if counter > 0:
                while pos[0] > 0 + counter:
                    if not update_results(): # no need to keep traversing
                        return results
                    move_north()
                    traversed = True
            while pos[1] < n - counter - 1:
                if not update_results(): # no need to keep traversing
                    return results
                move_east()
                traversed = True
            while pos[0] < m - counter - 1:
                if not update_results(): # no need to keep traversing
                    return results
                move_south()
                traversed = True
            while pos[1] > 0 + counter:
                if not update_results(): # no need to keep traversing
                    return results
                move_west()
                traversed = True

            if not traversed: #if there is the same number of elements in head as m*n, this will input the last element
                update_results()    
            counter += 1

        return results

        



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