#2304 - Minimum Path Cost in a Grid
Problem Description
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.
Solution
/**
* @param {number[][]} grid
* @param {number[][]} moveCost
* @return {number}
*/
var minPathCost = function(grid, moveCost) {
const m = grid.length;
const n = grid[0].length;
const dp = Array.from({ length: m }, () => new Array(n).fill(Infinity));
for (let j = 0; j < n; j++) {
dp[0][j] = grid[0][j];
}
for (let i = 0; i < m - 1; i++) {
for (let j = 0; j < n; j++) {
const value = grid[i][j];
for (let k = 0; k < n; k++) {
dp[i + 1][k] = Math.min(
dp[i + 1][k],
dp[i][j] + grid[i + 1][k] + moveCost[value][k]
);
}
}
}
return Math.min(...dp[m - 1]);
};