leetcode-problemset/leetcode-cn/originData/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.json

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            "title": "Minimum Number of Flips to Convert Binary Matrix to Zero Matrix",
            "titleSlug": "minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix",
            "content": "<p>Given a <code>m x n</code> binary matrix <code>mat</code>. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing <code>1</code> to <code>0</code> and <code>0</code> to <code>1</code>). A pair of cells are called neighbors if they share one edge.</p>\n\n<p>Return the <em>minimum number of steps</em> required to convert <code>mat</code> to a zero matrix or <code>-1</code> if you cannot.</p>\n\n<p>A <strong>binary matrix</strong> is a matrix with all cells equal to <code>0</code> or <code>1</code> only.</p>\n\n<p>A <strong>zero matrix</strong> is a matrix with all cells equal to <code>0</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/11/28/matrix.png\" style=\"width: 409px; height: 86px;\" />\n<pre>\n<strong>Input:</strong> mat = [[0,0],[0,1]]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> mat = [[0]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Given matrix is a zero matrix. We do not need to change it.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> mat = [[1,0,0],[1,0,0]]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> Given matrix cannot be a zero matrix.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == mat.length</code></li>\n\t<li><code>n == mat[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 3</code></li>\n\t<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>\n</ul>\n",
            "translatedTitle": "转化为全零矩阵的最少反转次数",
            "translatedContent": "<p>给你一个&nbsp;<code>m x n</code>&nbsp;的二进制矩阵&nbsp;<code>mat</code>。每一步，你可以选择一个单元格并将它反转（反转表示 <code>0</code> 变 <code>1</code> ，<code>1</code> 变 <code>0</code> ）。如果存在和它相邻的单元格，那么这些相邻的单元格也会被反转。相邻的两个单元格共享同一条边。</p>\n\n<p>请你返回将矩阵&nbsp;<code>mat</code> 转化为全零矩阵的<em>最少反转次数</em>，如果无法转化为全零矩阵，请返回&nbsp;<code>-1</code>&nbsp;。</p>\n\n<p><strong>二进制矩阵</strong>&nbsp;的每一个格子要么是 <code>0</code> 要么是 <code>1</code> 。</p>\n\n<p><strong>全零矩阵</strong>&nbsp;是所有格子都为 <code>0</code> 的矩阵。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例&nbsp;1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/12/13/matrix.png\" /></p>\n\n<pre>\n<strong>输入：</strong>mat = [[0,0],[0,1]]\n<strong>输出：</strong>3\n<strong>解释：</strong>一个可能的解是反转 (1, 0)，然后 (0, 1) ，最后是 (1, 1) 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>mat = [[0]]\n<strong>输出：</strong>0\n<strong>解释：</strong>给出的矩阵是全零矩阵，所以你不需要改变它。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>mat = [[1,0,0],[1,0,0]]\n<strong>输出：</strong>-1\n<strong>解释：</strong>该矩阵无法转变成全零矩阵\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m ==&nbsp;mat.length</code></li>\n\t<li><code>n ==&nbsp;mat[0].length</code></li>\n\t<li><code>1 &lt;= m&nbsp;&lt;= 3</code></li>\n\t<li><code>1 &lt;= n&nbsp;&lt;= 3</code></li>\n\t<li><code>mat[i][j]</code>&nbsp;是 0 或 1 。</li>\n</ul>\n",
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