mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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192 lines
19 KiB
JSON
192 lines
19 KiB
JSON
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"questionId": "2101",
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"title": "Last Day Where You Can Still Cross",
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"content": "<p>There is a <strong>1-based</strong> binary matrix where <code>0</code> represents land and <code>1</code> represents water. You are given integers <code>row</code> and <code>col</code> representing the number of rows and columns in the matrix, respectively.</p>\n\n<p>Initially on day <code>0</code>, the <strong>entire</strong> matrix is <strong>land</strong>. However, each day a new cell becomes flooded with <strong>water</strong>. You are given a <strong>1-based</strong> 2D array <code>cells</code>, where <code>cells[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> represents that on the <code>i<sup>th</sup></code> day, the cell on the <code>r<sub>i</sub><sup>th</sup></code> row and <code>c<sub>i</sub><sup>th</sup></code> column (<strong>1-based</strong> coordinates) will be covered with <strong>water</strong> (i.e., changed to <code>1</code>).</p>\n\n<p>You want to find the <strong>last</strong> day that it is possible to walk from the <strong>top</strong> to the <strong>bottom</strong> by only walking on land cells. You can start from <strong>any</strong> cell in the top row and end at <strong>any</strong> cell in the bottom row. You can only travel in the<strong> four</strong> cardinal directions (left, right, up, and down).</p>\n\n<p>Return <em>the <strong>last</strong> day where it is possible to walk from the <strong>top</strong> to the <strong>bottom</strong> by only walking on land cells</em>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/27/1.png\" style=\"width: 624px; height: 162px;\" />\n<pre>\n<strong>Input:</strong> row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.\nThe last day where it is possible to cross from top to bottom is on day 2.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/27/2.png\" style=\"width: 504px; height: 178px;\" />\n<pre>\n<strong>Input:</strong> row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.\nThe last day where it is possible to cross from top to bottom is on day 1.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/27/3.png\" style=\"width: 666px; height: 167px;\" />\n<pre>\n<strong>Input:</strong> row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> The above image depicts how the matrix changes each day starting from day 0.\nThe last day where it is possible to cross from top to bottom is on day 3.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= row, col <= 2 * 10<sup>4</sup></code></li>\n\t<li><code>4 <= row * col <= 2 * 10<sup>4</sup></code></li>\n\t<li><code>cells.length == row * col</code></li>\n\t<li><code>1 <= r<sub>i</sub> <= row</code></li>\n\t<li><code>1 <= c<sub>i</sub> <= col</code></li>\n\t<li>All the values of <code>cells</code> are <strong>unique</strong>.</li>\n</ul>\n",
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"code": "# @param {Integer} row\n# @param {Integer} col\n# @param {Integer[][]} cells\n# @return {Integer}\ndef latest_day_to_cross(row, col, cells)\n \nend",
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"code": "impl Solution {\n pub fn latest_day_to_cross(row: i32, col: i32, cells: Vec<Vec<i32>>) -> i32 {\n \n }\n}",
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"code": "class Solution {\n\n /**\n * @param Integer $row\n * @param Integer $col\n * @param Integer[][] $cells\n * @return Integer\n */\n function latestDayToCross($row, $col, $cells) {\n \n }\n}",
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"What graph algorithm allows us to find whether a path exists?",
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