# 2304. Minimum Path Cost in a Grid

2304. Minimum Path Cost in a Grid

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    Difficulty: Medium

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to 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.

 

Example 1:

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.

Example 2:

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.

 

Constraints:

    m == grid.length
    n == grid[i].length
    2 <= m, n <= 50
    grid consists of distinct integers from 0 to m * n - 1.
    moveCost.length == m * n
    moveCost[i].length == n
    1 <= moveCost[i][j] <= 100

class Solution:
    def minPathCost(self, grid: List[List[int]], moveCost: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])
        
        grid2 = [[0] * n for i in range(m) ] 
        # to save the min cost to arrive grid[i][j] from arbitrary starting point
        
        for i in range(1,m):  # skip 0th row
            for j in range(n):  
            # iter over every col, will jump to grid[i][j] from previous row
                grid2[i][j] += min(  
                      grid2[i-1][k] 
                      + grid[i-1][k] 
                      + moveCost[ grid[i-1][k] ][ j ] 
                  for k in range(n)  
                )
                # the min cost to arrive grid[i][j] is: the min among the sum: 
                #     grid2[i-1][k]                 :  the min cost to arrive the previous cell
                #   + grid[i-1][k]                  :  the value in previous cell 
                #   + moveCost[ grid[i-1][k] ][ j ] : the cost from previous cell to this grid[i][j] 
                #     
        return min(grid[m-1][i] + grid2[m-1][i]  for i in range(n))
        # ans is the min of (cost to arrive + value in cell)