Medium
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.
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 <= 105
0 <= rowCosts[r], colCosts[c] <= 104
startPos.length == 2
homePos.length == 2
0 <= startrow, homerow < m
0 <= startcol, homecol < n
class Solution {
fun minCost(startPos: IntArray, homePos: IntArray, rowCosts: IntArray, colCosts: IntArray): Int {
val sx = startPos[0]
val sy = startPos[1]
val ex = homePos[0]
val ey = homePos[1]
if (sx == ex && sy == ey) {
return 0
}
var total = 0
if (sx < ex) {
var i = sx
while (i < ex) {
i++
total += rowCosts[i]
}
} else {
var i = sx
while (i > ex) {
i--
total += rowCosts[i]
}
}
if (sy < ey) {
var i = sy
while (i < ey) {
i++
total += colCosts[i]
}
} else {
var i = sy
while (i > ey) {
i--
total += colCosts[i]
}
}
return total
}
}