leetcode-problemset/leetcode-cn/originData/flip-columns-for-maximum-number-of-equal-rows.json

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            "questionId": "1147",
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            "categoryTitle": "Algorithms",
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            "title": "Flip Columns For Maximum Number of Equal Rows",
            "titleSlug": "flip-columns-for-maximum-number-of-equal-rows",
            "content": "<p>You are given an <code>m x n</code> binary matrix <code>matrix</code>.</p>\n\n<p>You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from <code>0</code> to <code>1</code> or vice versa).</p>\n\n<p>Return <em>the maximum number of rows that have all values equal after some number of flips</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> matrix = [[0,1],[1,1]]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> After flipping no values, 1 row has all values equal.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> matrix = [[0,1],[1,0]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> After flipping values in the first column, both rows have equal values.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> matrix = [[0,0,0],[0,0,1],[1,1,0]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> After flipping values in the first two columns, the last two rows have equal values.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 300</code></li>\n\t<li><code>matrix[i][j]</code> is either&nbsp;<code>0</code> or <code>1</code>.</li>\n</ul>\n",
            "translatedTitle": "按列翻转得到最大值等行数",
            "translatedContent": "<p>给定&nbsp;<code>m x n</code>&nbsp;矩阵&nbsp;<code>matrix</code>&nbsp;。</p>\n\n<p>你可以从中选出任意数量的列并翻转其上的&nbsp;<strong>每个&nbsp;</strong>单元格。（即翻转后，单元格的值从 <code>0</code> 变成 <code>1</code>，或者从 <code>1</code> 变为 <code>0</code> 。）</p>\n\n<p>返回 <em>经过一些翻转后，行与行之间所有值都相等的最大行数</em>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<ol>\n</ol>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>matrix = [[0,1],[1,1]]\n<strong>输出：</strong>1\n<strong>解释：</strong>不进行翻转，有 1 行所有值都相等。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>matrix = [[0,1],[1,0]]\n<strong>输出：</strong>2\n<strong>解释：</strong>翻转第一列的值之后，这两行都由相等的值组成。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>matrix = [[0,0,0],[0,0,1],[1,1,0]]\n<strong>输出：</strong>2\n<strong>解释：</strong>翻转前两列的值之后，后两行由相等的值组成。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 300</code></li>\n\t<li><code>matrix[i][j] == 0</code> 或&nbsp;<code>1</code></li>\n</ul>\n",
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                "Flipping a subset of columns is like doing a bitwise XOR of some number K onto each row.  We want rows X with X ^ K = all 0s or all 1s.  This is the same as X = X^K ^K = (all 0s or all 1s) ^ K, so we want to count rows that have opposite bits set.  For example, if K = 1, then we count rows X = (00000...001, or 1111....110)."
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