<p>Given an <code>m x n</code> <code>matrix</code>, return <em>all elements of the</em> <code>matrix</code> <em>in spiral order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,3,6,9,8,7,4,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
<strong>Output:</strong> [1,2,3,4,8,12,11,10,9,5,6,7]
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
</ul>

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        top, left = 0, 0
        bottom = m - 1
        right = n - 1
        
        output = []

        while left <= right and top <= bottom:

            # top row
            for i in range(left, right + 1):
                output.append(matrix[top][i])

            # right col
            for i in range(top + 1, bottom + 1):
                output.append(matrix[i][right])

            # bottom row
            if top != bottom:
                for i in range(right - 1, left - 1, -1):
                    output.append(matrix[bottom][i])

            # left col
            if right != left:

                for i in range(bottom - 1, top, -1):
                    output.append(matrix[i][left])
            left += 1
            right -= 1
            top += 1
            bottom -= 1
        
        return output
            

        return output
16500