2304. Minimum Path Cost in a Grid

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.

class Solution {
public:
    int minPathCost(vector<vector<int>>& grid, vector<vector<int>>& moveCost) {
        int n = grid.size(), m = grid[0].size();
        vector<int> dp = grid[0];
        for(int i = 1; i < n; i++) {
            vector<int> ndp(m,INT_MAX);
            for(int j = 0; j < m; j++) {
                auto& c = moveCost[grid[i-1][j]];
                for(int k = 0; k < m; k++) {
                    ndp[k] = min(ndp[k], dp[j] + grid[i][k] + c[k]);
                }
            }
            swap(dp, ndp);
        }
        return *min_element(begin(dp), end(dp));
    }
};