leetcode-problemset/leetcode-cn/originData/difference-of-number-of-distinct-values-on-diagonals.json

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        "question": {
            "questionId": "2801",
            "questionFrontendId": "2711",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2284417,
            "title": "Difference of Number of Distinct Values on Diagonals",
            "titleSlug": "difference-of-number-of-distinct-values-on-diagonals",
            "content": "<p>Given a 2D <code>grid</code> of size <code>m x n</code>, you should find the matrix <code>answer</code> of size <code>m x n</code>.</p>\n\n<p>The cell <code>answer[r][c]</code> is calculated by looking at the diagonal values of the cell <code>grid[r][c]</code>:</p>\n\n<ul>\n\t<li>Let <code>leftAbove[r][c]</code> be the number of <strong>distinct</strong> values on the diagonal to the left and above the cell <code>grid[r][c]</code> not including the cell <code>grid[r][c]</code> itself.</li>\n\t<li>Let <code>rightBelow[r][c]</code> be the number of <strong>distinct</strong> values on the diagonal to the right and below the cell <code>grid[r][c]</code>, not including the cell <code>grid[r][c]</code> itself.</li>\n\t<li>Then <code>answer[r][c] = |leftAbove[r][c] - rightBelow[r][c]|</code>.</li>\n</ul>\n\n<p>A <strong>matrix diagonal</strong> is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until the end of the matrix is reached.</p>\n\n<ul>\n\t<li>For example, in the below diagram the diagonal is highlighted using the cell with indices <code>(2, 3)</code> colored gray:\n\n\t<ul>\n\t\t<li>Red-colored cells are left and above the cell.</li>\n\t\t<li>Blue-colored cells are right and below the cell.</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/05/26/diagonal.png\" style=\"width: 200px; height: 160px;\" /></p>\n\n<p>Return the matrix <code>answer</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">grid = [[1,2,3],[3,1,5],[3,2,1]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">Output: [[1,1,0],[1,0,1],[0,1,1]]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>To calculate the <code>answer</code> cells:</p>\n\n<table>\n\t<thead>\n\t\t<tr>\n\t\t\t<th>answer</th>\n\t\t\t<th>left-above elements</th>\n\t\t\t<th>leftAbove</th>\n\t\t\t<th>right-below elements</th>\n\t\t\t<th>rightBelow</th>\n\t\t\t<th>|leftAbove - rightBelow|</th>\n\t\t</tr>\n\t</thead>\n\t<tbody>\n\t\t<tr>\n\t\t\t<td>[0][0]</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>[grid[1][1], grid[2][2]]</td>\n\t\t\t<td>|{1, 1}| = 1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0][1]</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>[grid[1][2]]</td>\n\t\t\t<td>|{5}| = 1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0][2]</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[1][0]</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>[grid[2][1]]</td>\n\t\t\t<td>|{2}| = 1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[1][1]</td>\n\t\t\t<td>[grid[0][0]]</td>\n\t\t\t<td>|{1}| = 1</td>\n\t\t\t<td>[grid[2][2]]</td>\n\t\t\t<td>|{1}| = 1</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[1][2]</td>\n\t\t\t<td>[grid[0][1]]</td>\n\t\t\t<td>|{2}| = 1</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2][0]</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2][1]</td>\n\t\t\t<td>[grid[1][0]]</td>\n\t\t\t<td>|{3}| = 1</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2][2]</td>\n\t\t\t<td>[grid[0][0], grid[1][1]]</td>\n\t\t\t<td>|{1, 1}| = 1</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">grid = [[1]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">Output: [[0]]</span></p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>1 &lt;= m, n, grid[i][j] &lt;= 50</code></li>\n</ul>\n",
            "translatedTitle": "对角线上不同值的数量差",
            "translatedContent": "<p>给你一个下标从 <code>0</code> 开始、大小为 <code>m x n</code> 的二维矩阵 <code>grid</code> ，请你求解大小同样为 <code>m x n</code> 的答案矩阵 <code>answer</code> 。</p>\n\n<p>矩阵 <code>answer</code> 中每个单元格 <code>(r, c)</code> 的值可以按下述方式进行计算：</p>\n\n<ul>\n\t<li>令 <code>topLeft[r][c]</code> 为矩阵 <code>grid</code> 中单元格 <code>(r, c)</code> 左上角对角线上 <strong>不同值</strong> 的数量。</li>\n\t<li>令 <code>bottomRight[r][c]</code> 为矩阵 <code>grid</code> 中单元格 <code>(r, c)</code> 右下角对角线上 <strong>不同值</strong> 的数量。</li>\n</ul>\n\n<p>然后 <code>answer[r][c] = |topLeft[r][c] - bottomRight[r][c]|</code> 。</p>\n\n<p>返回矩阵 <code>answer</code> 。</p>\n\n<p><strong>矩阵对角线</strong> 是从最顶行或最左列的某个单元格开始，向右下方向走到矩阵末尾的对角线。</p>\n\n<p>如果单元格 <code>(r1, c1)</code> 和单元格 <code>(r, c) </code>属于同一条对角线且 <code>r1 &lt; r</code> ，则单元格 <code>(r1, c1)</code> 属于单元格 <code>(r, c)</code> 的左上对角线。类似地，可以定义右下对角线。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/04/19/ex2.png\" style=\"width: 786px; height: 121px;\" />\n<pre>\n<strong>输入：</strong>grid = [[1,2,3],[3,1,5],[3,2,1]]\n<strong>输出：</strong>[[1,1,0],[1,0,1],[0,1,1]]\n<strong>解释：</strong>第 1 个图表示最初的矩阵 grid 。&nbsp;\n第 2 个图表示对单元格 (0,0) 计算，其中蓝色单元格是位于右下对角线的单元格。\n第 3 个图表示对单元格 (1,2) 计算，其中红色单元格是位于左上对角线的单元格。\n第 4 个图表示对单元格 (1,1) 计算，其中蓝色单元格是位于右下对角线的单元格，红色单元格是位于左上对角线的单元格。\n- 单元格 (0,0) 的右下对角线包含 [1,1] ，而左上对角线包含 [] 。对应答案是 |1 - 0| = 1 。\n- 单元格 (1,2) 的右下对角线包含 [] ，而左上对角线包含 [2] 。对应答案是 |0 - 1| = 1 。\n- 单元格 (1,1) 的右下对角线包含 [1] ，而左上对角线包含 [1] 。对应答案是 |1 - 1| = 0 。\n其他单元格的对应答案也可以按照这样的流程进行计算。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>grid = [[1]]\n<strong>输出：</strong>[[0]]\n<strong>解释：</strong>- 单元格 (0,0) 的右下对角线包含 [] ，左上对角线包含 [] 。对应答案是 |0 - 0| = 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>1 &lt;= m, n, grid[i][j] &lt;= 50</code></li>\n</ul>\n",
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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