You are given a 0-indexed
m x n integer matrix
grid consisting of distinct integers from
m * n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell
(x, y) such that
x < m - 1, you can move to any of the cells
(x + 1, 0),
(x + 1, 1), ...,
(x + 1, n - 1). Note that it is not possible to move from cells in the last row.
Each possible move has a cost given by a 0-indexed 2D array
moveCost of size
(m * n) x n, where
moveCost[i][j] is the cost of moving from a cell with value
i to a cell in column
j of the next row. The cost of moving from cells in the last row of
grid can be ignored.
The cost of a path in
grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.
Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]] Output: 17 Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1. - The sum of the values of cells visited is 5 + 0 + 1 = 6. - The cost of moving from 5 to 0 is 3. - The cost of moving from 0 to 1 is 8. So the total cost of the path is 6 + 3 + 8 = 17.
Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]] Output: 6 Explanation: The path with the minimum possible cost is the path 2 -> 3. - The sum of the values of cells visited is 2 + 3 = 5. - The cost of moving from 2 to 3 is 1. So the total cost of this path is 5 + 1 = 6.
m == grid.length
n == grid[i].length
2 <= m, n <= 50
gridconsists of distinct integers from
m * n - 1.
moveCost.length == m * n
moveCost[i].length == n
1 <= moveCost[i][j] <= 100