leetcode-problemset/leetcode-cn/originData/largest-plus-sign.json

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        "question": {
            "questionId": "769",
            "questionFrontendId": "764",
            "categoryTitle": "Algorithms",
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            "title": "Largest Plus Sign",
            "titleSlug": "largest-plus-sign",
            "content": "<p>You are given an integer <code>n</code>. You have an <code>n x n</code> binary grid <code>grid</code> with all values initially <code>1</code>&#39;s except for some indices given in the array <code>mines</code>. The <code>i<sup>th</sup></code> element of the array <code>mines</code> is defined as <code>mines[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> where <code>grid[x<sub>i</sub>][y<sub>i</sub>] == 0</code>.</p>\n\n<p>Return <em>the order of the largest <strong>axis-aligned</strong> plus sign of </em>1<em>&#39;s contained in </em><code>grid</code>. If there is none, return <code>0</code>.</p>\n\n<p>An <strong>axis-aligned plus sign</strong> of <code>1</code>&#39;s of order <code>k</code> has some center <code>grid[r][c] == 1</code> along with four arms of length <code>k - 1</code> going up, down, left, and right, and made of <code>1</code>&#39;s. Note that there could be <code>0</code>&#39;s or <code>1</code>&#39;s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for <code>1</code>&#39;s.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/13/plus1-grid.jpg\" style=\"width: 404px; height: 405px;\" />\n<pre>\n<strong>Input:</strong> n = 5, mines = [[4,2]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> In the above grid, the largest plus sign can only be of order 2. One of them is shown.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/13/plus2-grid.jpg\" style=\"width: 84px; height: 85px;\" />\n<pre>\n<strong>Input:</strong> n = 1, mines = [[0,0]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> There is no plus sign, so return 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= mines.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt; n</code></li>\n\t<li>All the pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are <strong>unique</strong>.</li>\n</ul>\n",
            "translatedTitle": "最大加号标志",
            "translatedContent": "<p>在一个 <code>n x n</code> 的矩阵&nbsp;<code>grid</code>&nbsp;中，除了在数组&nbsp;<code>mines</code>&nbsp;中给出的元素为&nbsp;<code>0</code>，其他每个元素都为&nbsp;<code>1</code>。<code>mines[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>表示&nbsp;<code>grid[x<sub>i</sub>][y<sub>i</sub>] == 0</code></p>\n\n<p>返回 <em>&nbsp;</em><code>grid</code><em> 中包含&nbsp;<code>1</code>&nbsp;的最大的 <strong>轴对齐</strong> 加号标志的阶数</em> 。如果未找到加号标志，则返回 <code>0</code> 。</p>\n\n<p>一个&nbsp;<code>k</code>&nbsp;阶由&nbsp;<em><code>1</code></em>&nbsp;组成的 <strong>“轴对称”加号标志</strong> 具有中心网格&nbsp;<code>grid[r][c] == 1</code>&nbsp;，以及4个从中心向上、向下、向左、向右延伸，长度为&nbsp;<code>k-1</code>，由&nbsp;<code>1</code>&nbsp;组成的臂。注意，只有加号标志的所有网格要求为 <code>1</code> ，别的网格可能为 <code>0</code> 也可能为 <code>1</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2021/06/13/plus1-grid.jpg\" /></p>\n\n<pre>\n<strong>输入:</strong> n = 5, mines = [[4, 2]]\n<strong>输出:</strong> 2\n<strong>解释: </strong>在上面的网格中，最大加号标志的阶只能是2。一个标志已在图中标出。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2021/06/13/plus2-grid.jpg\" /></p>\n\n<pre>\n<strong>输入:</strong> n = 1, mines = [[0, 0]]\n<strong>输出:</strong> 0\n<strong>解释: </strong>没有加号标志，返回 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= mines.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub>&nbsp;&lt; n</code></li>\n\t<li>每一对&nbsp;<code>(x<sub>i</sub>, y<sub>i</sub>)</code>&nbsp;都 <strong>不重复</strong>​​​​​​​</li>\n</ul>\n",
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                "For each direction such as \"left\", find left[r][c] = the number of 1s you will see before a zero starting at r, c and walking left.  You can find this in N^2 time with a dp.  The largest plus sign at r, c is just the minimum of left[r][c], up[r][c] etc."
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