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#1368 - Minimum Cost to Make at Least One Valid Path in a Grid
Problem Description
Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:
- 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
- 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
- 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
- 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])
Notice that there could be some signs on the cells of the grid that point outside the grid.
You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.
You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.
Return the minimum cost to make the grid have at least one valid path.
Solution
/**
* @param {number[][]} grid
* @return {number}
*/
var minCost = function(grid) {
const queue = [[0, 0]];
const directions = [[0, 1, 1], [0, -1, 2], [1, 0, 3], [-1, 0, 4]];
const bfs = grid.map(r => r.map(_ => Infinity));
bfs[0][0] = 0;
while (queue.length > 0) {
const [x, y] = queue.shift();
for (const [dx, dy, value] of directions) {
const [cX, cY] = [x + dx, y + dy];
if (grid[cX]?.[cY]) {
const updatedValue = bfs[x][y] + (grid[x][y] !== value);
if (updatedValue < bfs[cX][cY]) {
bfs[cX][cY] = updatedValue;
queue.push([cX, cY]);
}
}
}
}
return bfs.at(-1).at(-1);
};