2087. Minimum Cost Homecoming of a Robot in a Grid

Problem

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.

• 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[2] = 3.
-> It goes right to (2, 1). This move costs colCosts[1] = 2.
-> It goes right to (2, 2). This move costs colCosts[2] = 6.
-> It goes right to (2, 3). This move costs colCosts[3] = 7.
The total cost is 3 + 2 + 6 + 7 = 18

Example 2:

Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]
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 <= 10^5

• 0 <= rowCosts[r], colCosts[c] <= 10^4

• startPos.length == 2

• homePos.length == 2

• 0 <= startrow, homerow < m

• 0 <= startcol, homecol < n

Solution (Java)

class Solution {
    public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
        int cost = 0;
        for (int i = Math.min(startPos[0], homePos[0]);
                i <= Math.max(startPos[0], homePos[0]);
                i++) {
            cost += rowCosts[i];
        }
        for (int j = Math.min(startPos[1], homePos[1]);
                j <= Math.max(startPos[1], homePos[1]);
                j++) {
            cost += colCosts[j];
        }
        return cost - rowCosts[startPos[0]] - colCosts[startPos[1]];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).