<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>'s.</p> <p>You must do it <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in place</a>.</p> <p> </p> <p><strong>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>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> </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 <= m, n <= 200</code></li> <li><code>-2<sup>31</sup> <= matrix[i][j] <= 2<sup>31</sup> - 1</code></li> </ul> <p> </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>