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leetcode/problem/score-after-flipping-matrix.html
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leetcode/problem/score-after-flipping-matrix.html
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<p>You are given an <code>m x n</code> binary matrix <code>grid</code>.</p>
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<p>A <strong>move</strong> consists of choosing any row or column and toggling each value in that row or column (i.e., changing all <code>0</code>'s to <code>1</code>'s, and all <code>1</code>'s to <code>0</code>'s).</p>
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<p>Every row of the matrix is interpreted as a binary number, and the <strong>score</strong> of the matrix is the sum of these numbers.</p>
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<p>Return <em>the highest possible <strong>score</strong> after making any number of <strong>moves</strong> (including zero moves)</em>.</p>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-toogle1.jpg" style="width: 500px; height: 299px;" />
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<pre>
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<strong>Input:</strong> grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
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<strong>Output:</strong> 39
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<strong>Explanation:</strong> 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> grid = [[0]]
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<strong>Output:</strong> 1
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>m == grid.length</code></li>
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<li><code>n == grid[i].length</code></li>
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<li><code>1 <= m, n <= 20</code></li>
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<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
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</ul>
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