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42 lines
1.8 KiB
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42 lines
1.8 KiB
HTML
<p>You are given a <strong>0-indexed</strong> <code>m x n</code> binary matrix <code>mat</code> and an integer <code>cols</code>, which denotes the number of columns you must choose.</p>
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<p>A row is <strong>covered</strong> by a set of columns if each cell in the row that has a value of <code>1</code> also lies in one of the columns of the chosen set.</p>
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<p>Return <em>the <strong>maximum</strong> number of rows that can be <strong>covered</strong> by a set of </em><code>cols</code><em> columns.</em></p>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered.png" style="width: 250px; height: 417px;" /></strong></p>
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<pre>
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<strong>Input:</strong> mat = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], cols = 2
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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As shown in the diagram above, one possible way of covering 3 rows is by selecting the 0th and 2nd columns.
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It can be shown that no more than 3 rows can be covered, so we return 3.
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</pre>
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<p><strong>Example 2:</strong></p>
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<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered2.png" style="width: 83px; height: 247px;" /></strong></p>
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<pre>
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<strong>Input:</strong> mat = [[1],[0]], cols = 1
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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Selecting the only column will result in both rows being covered, since the entire matrix is selected.
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Therefore, we return 2.
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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 == mat.length</code></li>
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<li><code>n == mat[i].length</code></li>
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<li><code>1 <= m, n <= 12</code></li>
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<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
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<li><code>1 <= cols <= n</code></li>
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</ul>
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