<p>Given an <code>m x n</code> integer matrix <code>matrix</code>, if an element is <code>0</code>, set its entire row and column to <code>0</code>&#39;s.</p>

<p>You must do it <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in place</a>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg" style="width: 450px; height: 169px;" />
<pre>
<strong>Input:</strong> matrix = [[1,1,1],[1,0,1],[1,1,1]]
<strong>Output:</strong> [[1,0,1],[0,0,0],[1,0,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg" style="width: 450px; height: 137px;" />
<pre>
<strong>Input:</strong> matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
<strong>Output:</strong> [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[0].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>-2<sup>31</sup> &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>A straightforward solution using <code>O(mn)</code> space is probably a bad idea.</li>
	<li>A simple improvement uses <code>O(m + n)</code> space, but still not the best solution.</li>
	<li>Could you devise a constant space solution?</li>
</ul>

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        zero_col = set()
        zero_row = set()

        m = len(matrix)
        n = len(matrix[0])

        for i in range(m):
            for j in range(n):
                if matrix[i][j] == 0:
                    zero_col.add(j)
                    zero_row.add(i)

        for i in range(m):
            if i in zero_row:
                matrix[i] = [0] * n
                continue
            for j in range(n):
                if j in zero_col:
                    matrix[i][j] = 0
        
        return matrix
17200