There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates that initially, a robot is at the cell (startrow, startcol). You are also given an integer array homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, homecol).
The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n.
r, then this move costs rowCosts[r].c, then this move costs colCosts[c].Return the minimum total cost for this robot to return home.
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Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2021/10/11/eg-1.png\")
\n\nInput: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]\nOutput: 18\nExplanation: One optimal path is that:\nStarting from (1, 0)\n-> It goes down to (2, 0). This move costs rowCosts[2] = 3.\n-> It goes right to (2, 1). This move costs colCosts[1] = 2.\n-> It goes right to (2, 2). This move costs colCosts[2] = 6.\n-> It goes right to (2, 3). This move costs colCosts[3] = 7.\nThe total cost is 3 + 2 + 6 + 7 = 18\n\n
Example 2:
\n\n\nInput: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]\nOutput: 0\nExplanation: The robot is already at its home. Since no moves occur, the total cost is 0.\n\n\n
\n
Constraints:
\n\nm == rowCosts.lengthn == colCosts.length1 <= m, n <= 1050 <= rowCosts[r], colCosts[c] <= 104startPos.length == 2homePos.length == 20 <= startrow, homerow < m0 <= startcol, homecol < nCompiled with clang 11 using the latest C++ 17 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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