#3603

Minimum Cost Path with Alternating Directions II

specialist · 960 · lc medium +32 · 44.6% accepted · 72 likes · top 27%

Description

A grid has m rows and n columns. Entering cell (i, j) costs (i + 1) * (j + 1). A second array waitCost[i][j] gives the cost of waiting at that cell.

You enter (0, 0) on second 1, paying its entry cost. Movement alternates:

- On odd seconds: move right or down to an adjacent cell and pay its entry cost.

- On even seconds: remain in place for one second and pay waitCost[i][j].

Return the minimum total cost to reach (m - 1, n - 1).

Code

1

2

3

class Solution:
    def minCost(self, m: int, n: int, waitCost: List[List[int]]) -> int:

class Solution:
    def minCost(self, m: int, n: int, waitCost: List[List[int]]) -> int: