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            "title": "Minimum Cost Homecoming of a Robot in a Grid",
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            "content": "<p>There is an <code>m x n</code> grid, where <code>(0, 0)</code> is the top-left cell and <code>(m - 1, n - 1)</code> is the bottom-right cell. You are given an integer array <code>startPos</code> where <code>startPos = [start<sub>row</sub>, start<sub>col</sub>]</code> indicates that <strong>initially</strong>, a <strong>robot</strong> is at the cell <code>(start<sub>row</sub>, start<sub>col</sub>)</code>. You are also given an integer array <code>homePos</code> where <code>homePos = [home<sub>row</sub>, home<sub>col</sub>]</code> indicates that its <strong>home</strong> is at the cell <code>(home<sub>row</sub>, home<sub>col</sub>)</code>.</p>\n\n<p>The robot needs to go to its home. It can move one cell in four directions: <strong>left</strong>, <strong>right</strong>, <strong>up</strong>, or <strong>down</strong>, and it can not move outside the boundary. Every move incurs some cost. You are further given two <strong>0-indexed</strong> integer arrays: <code>rowCosts</code> of length <code>m</code> and <code>colCosts</code> of length <code>n</code>.</p>\n\n<ul>\n\t<li>If the robot moves <strong>up</strong> or <strong>down</strong> into a cell whose <strong>row</strong> is <code>r</code>, then this move costs <code>rowCosts[r]</code>.</li>\n\t<li>If the robot moves <strong>left</strong> or <strong>right</strong> into a cell whose <strong>column</strong> is <code>c</code>, then this move costs <code>colCosts[c]</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum total cost</strong> for this robot to return home</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/10/11/eg-1.png\" style=\"width: 282px; height: 217px;\" />\n<pre>\n<strong>Input:</strong> startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]\n<strong>Output:</strong> 18\n<strong>Explanation:</strong> One optimal path is that:\nStarting from (1, 0)\n-&gt; It goes down to (<u><strong>2</strong></u>, 0). This move costs rowCosts[2] = 3.\n-&gt; It goes right to (2, <u><strong>1</strong></u>). This move costs colCosts[1] = 2.\n-&gt; It goes right to (2, <u><strong>2</strong></u>). This move costs colCosts[2] = 6.\n-&gt; It goes right to (2, <u><strong>3</strong></u>). This move costs colCosts[3] = 7.\nThe total cost is 3 + 2 + 6 + 7 = 18</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The robot is already at its home. Since no moves occur, the total cost is 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == rowCosts.length</code></li>\n\t<li><code>n == colCosts.length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= rowCosts[r], colCosts[c] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>startPos.length == 2</code></li>\n\t<li><code>homePos.length == 2</code></li>\n\t<li><code>0 &lt;= start<sub>row</sub>, home<sub>row</sub> &lt; m</code></li>\n\t<li><code>0 &lt;= start<sub>col</sub>, home<sub>col</sub> &lt; n</code></li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int minCost(vector<int>& startPos, vector<int>& homePos, vector<int>& rowCosts, vector<int>& colCosts) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def minCost(self, startPos, homePos, rowCosts, colCosts):\n        \"\"\"\n        :type startPos: List[int]\n        :type homePos: List[int]\n        :type rowCosts: List[int]\n        :type colCosts: List[int]\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "/**\n * @param {number[]} startPos\n * @param {number[]} homePos\n * @param {number[]} rowCosts\n * @param {number[]} colCosts\n * @return {number}\n */\nvar minCost = function(startPos, homePos, rowCosts, colCosts) {\n    \n};",
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                    "code": "# @param {Integer[]} start_pos\n# @param {Integer[]} home_pos\n# @param {Integer[]} row_costs\n# @param {Integer[]} col_costs\n# @return {Integer}\ndef min_cost(start_pos, home_pos, row_costs, col_costs)\n    \nend",
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                    "code": "func minCost(startPos []int, homePos []int, rowCosts []int, colCosts []int) int {\n    \n}",
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                    "code": "object Solution {\n    def minCost(startPos: Array[Int], homePos: Array[Int], rowCosts: Array[Int], colCosts: Array[Int]): Int = {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun minCost(startPos: IntArray, homePos: IntArray, rowCosts: IntArray, colCosts: IntArray): Int {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn min_cost(start_pos: Vec<i32>, home_pos: Vec<i32>, row_costs: Vec<i32>, col_costs: Vec<i32>) -> i32 {\n        \n    }\n}",
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                    "code": "-spec min_cost(StartPos :: [integer()], HomePos :: [integer()], RowCosts :: [integer()], ColCosts :: [integer()]) -> integer().\nmin_cost(StartPos, HomePos, RowCosts, ColCosts) ->\n  .",
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                "Irrespective of what path the robot takes, it will have to traverse all the rows between startRow and homeRow and all the columns between startCol and homeCol.",
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