<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]]:
 
        #create the n * m matrix 
        output = [[None] * n for _ in range(m)]
        
        #create my top bottom left and right pointers 
        top = 0 
        bottom = m
        left = 0 
        right = n

        #While loop to keep iterating over stuff until boundaries overlap or something 
        while left < right and top < bottom: 
            #iterate over top row
            for col in range(left, right): 
                if head: 
                    print(head.val)
                    output[top][col] = head.val
                    head = head.next
                else: 
                    output[top][col] = -1

            top += 1

            #iterate over rightmost column 
            for row in range(top, bottom): 
                if head:
                    print(head.val)
                    output[row][right - 1] = head.val 
                    head = head.next
                else: 
                    output[row][right - 1] = -1
            right -= 1

            #check if bounds have changed 
            if not (left < right and top < bottom): 
                break

            #iterate over bottom row 
            for col in range(right - 1, left -1, -1): 
                if head: 
                    print(head.val)
                    output[bottom - 1][col] = head.val
                    head = head.next 
                else: 
                    output[bottom - 1][col] = -1

            bottom -= 1

            #iterate over left most column
            for row in range(bottom-1, top -1, -1): 
                if head: 
                    print(head.val)
                    output[row][left] = head.val 
                    head = head.next 
                else: 
                    output[row][left] = -1
            left += 1

        #return the filled in matrix 
        return output 

        
'''
understand: 
- Okay so I'm given the dimensions of the matrix, m rows and n columns, and a 
linked list with numbers. It's my job to fill the matrix with the values in the linked list in a 
clockwise spiral pattern 
- THe linked list might not have enough numbers, meaning i"ll have to fill some matrices with -1s the 
rest of the way 
- I can get a one dimensional matrix. list is max as long as there are spaces in the matrix 

match: 
- Okay it looks like I can probably use the 4 pointer boundary strategy to break this problem down 
into concentric circles basically
- I will have to create an empty n by m matrix first, at least thats the easiest thing for me to do 
- i think I can just keep adding numbers from the list and once I reach nullptr I can just fill the rest
with negative numbers 


'''
55000