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: leftrightup, 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.

• If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r].
• If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c].

Return the minimum total cost for this robot to return home.

Example 1:

Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]
Output: 18
Explanation: One optimal path is that:
Starting from (1, 0)
-> It goes down to (2, 0). This move costs rowCosts = 3.
-> It goes right to (2, 1). This move costs colCosts = 2.
-> It goes right to (2, 2). This move costs colCosts = 6.
-> It goes right to (2, 3). This move costs colCosts = 7.
The total cost is 3 + 2 + 6 + 7 = 18

Example 2:

Input: startPos = [0, 0], homePos = [0, 0], rowCosts = , colCosts = 
Output: 0
Explanation: The robot is already at its home. Since no moves occur, the total cost is 0.


Constraints:

• m == rowCosts.length
• n == colCosts.length
• 1 <= m, n <= 105
• 0 <= rowCosts[r], colCosts[c] <= 104
• startPos.length == 2
• homePos.length == 2
• 0 <= startrow, homerow < m
• 0 <= startcol, homecol < n

## Solution: Manhattan distance

Move directly to the goal, no back and forth. Cost will be the same no matter which path you choose.

ans = sum(rowCosts[y1+1~y2]) + sum(colCosts[x1+1~x2])

Time complexity: O(m + n)
Space complexity: O(1)

## C++

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