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43 lines
2.0 KiB
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<p>There is an <code>m x n</code> matrix that is initialized to all <code>0</code>'s. There is also a 2D array <code>indices</code> where each <code>indices[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> represents a <strong>0-indexed location</strong> to perform some increment operations on the matrix.</p>
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<p>For each location <code>indices[i]</code>, do <strong>both</strong> of the following:</p>
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<ol>
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<li>Increment <strong>all</strong> the cells on row <code>r<sub>i</sub></code>.</li>
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<li>Increment <strong>all</strong> the cells on column <code>c<sub>i</sub></code>.</li>
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</ol>
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<p>Given <code>m</code>, <code>n</code>, and <code>indices</code>, return <em>the <strong>number of odd-valued cells</strong> in the matrix after applying the increment to all locations in </em><code>indices</code>.</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/2019/10/30/e1.png" style="width: 600px; height: 118px;" />
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<pre>
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<strong>Input:</strong> m = 2, n = 3, indices = [[0,1],[1,1]]
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> Initial matrix = [[0,0,0],[0,0,0]].
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After applying first increment it becomes [[1,2,1],[0,1,0]].
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The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.
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</pre>
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<p><strong>Example 2:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e2.png" style="width: 600px; height: 150px;" />
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<pre>
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<strong>Input:</strong> m = 2, n = 2, indices = [[1,1],[0,0]]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.
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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>1 <= m, n <= 50</code></li>
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<li><code>1 <= indices.length <= 100</code></li>
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<li><code>0 <= r<sub>i</sub> < m</code></li>
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<li><code>0 <= c<sub>i</sub> < n</code></li>
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</ul>
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<p> </p>
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<p><strong>Follow up:</strong> Could you solve this in <code>O(n + m + indices.length)</code> time with only <code>O(n + m)</code> extra space?</p>
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