1391. Check if There is a Valid Path in a Grid

Problem

You are given an m x n grid. Each cell of grid represents a street. The street of grid[i][j] can be:

• 1 which means a street connecting the left cell and the right cell.

• 2 which means a street connecting the upper cell and the lower cell.

• 3 which means a street connecting the left cell and the lower cell.

• 4 which means a street connecting the right cell and the lower cell.

• 5 which means a street connecting the left cell and the upper cell.

• 6 which means a street connecting the right cell and the upper cell.

You will initially start at the street of 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). The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true** if there is a valid path in the grid or false otherwise**.

  Example 1:

Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

  Constraints:

• m == grid.length

• n == grid[i].length

• 1 <= m, n <= 300

• 1 <= grid[i][j] <= 6

Solution (Java)

class Solution {
    private int[][][] dirs = {
        {{0, -1}, {0, 1}},
        {{-1, 0}, {1, 0}},
        {{0, -1}, {1, 0}},
        {{0, 1}, {1, 0}},
        {{0, -1}, {-1, 0}},
        {{0, 1}, {-1, 0}}
    };
    // the idea is you need to check port direction match, you can go to next cell and check whether
    // you can come back.
    public boolean hasValidPath(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        boolean[][] visited = new boolean[m][n];
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[] {0, 0});
        visited[0][0] = true;
        while (!q.isEmpty()) {
            int[] cur = q.poll();
            int x = cur[0];
            int y = cur[1];
            int num = grid[x][y] - 1;
            for (int[] dir : dirs[num]) {
                int nx = x + dir[0];
                int ny = y + dir[1];
                if (nx < 0 || nx >= m || ny < 0 || ny >= n || visited[nx][ny]) {
                    continue;
                }
                // go to the next cell and come back to orign to see if port directions are same
                for (int[] backDir : dirs[grid[nx][ny] - 1]) {
                    if (nx + backDir[0] == x && ny + backDir[1] == y) {
                        visited[nx][ny] = true;
                        q.add(new int[] {nx, ny});
                    }
                }
            }
        }
        return visited[m - 1][n - 1];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).