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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>&nbsp;</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>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= row, col &lt;= 2 * 10<sup>4</sup></code></li>\n\t<li><code>4 &lt;= row * col &lt;= 2 * 10<sup>4</sup></code></li>\n\t<li><code>cells.length == row * col</code></li>\n\t<li><code>1 &lt;= r<sub>i</sub> &lt;= row</code></li>\n\t<li><code>1 &lt;= c<sub>i</sub> &lt;= 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": "class Solution(object):\n    def latestDayToCross(self, row, col, cells):\n        \"\"\"\n        :type row: int\n        :type col: int\n        :type cells: List[List[int]]\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "/**\n * @param {number} row\n * @param {number} col\n * @param {number[][]} cells\n * @return {number}\n */\nvar latestDayToCross = function(row, col, cells) {\n    \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": "object Solution {\n    def latestDayToCross(row: Int, col: Int, cells: Array[Array[Int]]): Int = {\n        \n    }\n}",
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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": "defmodule Solution do\n  @spec latest_day_to_cross(row :: integer, col :: integer, cells :: [[integer]]) :: integer\n  def latest_day_to_cross(row, col, cells) do\n\n  end\nend",
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