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

Description

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

Solutions

Solution 1

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class Solution:
    def hasValidPath(self, grid: List[List[int]]) -> bool:
        m, n = len(grid), len(grid[0])
        p = list(range(m * n))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def left(i, j):
            if j > 0 and grid[i][j - 1] in (1, 4, 6):
                p[find(i * n + j)] = find(i * n + j - 1)

        def right(i, j):
            if j < n - 1 and grid[i][j + 1] in (1, 3, 5):
                p[find(i * n + j)] = find(i * n + j + 1)

        def up(i, j):
            if i > 0 and grid[i - 1][j] in (2, 3, 4):
                p[find(i * n + j)] = find((i - 1) * n + j)

        def down(i, j):
            if i < m - 1 and grid[i + 1][j] in (2, 5, 6):
                p[find(i * n + j)] = find((i + 1) * n + j)

        for i in range(m):
            for j in range(n):
                e = grid[i][j]
                if e == 1:
                    left(i, j)
                    right(i, j)
                elif e == 2:
                    up(i, j)
                    down(i, j)
                elif e == 3:
                    left(i, j)
                    down(i, j)
                elif e == 4:
                    right(i, j)
                    down(i, j)
                elif e == 5:
                    left(i, j)
                    up(i, j)
                else:
                    right(i, j)
                    up(i, j)
        return find(0) == find(m * n - 1)

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class Solution {
    private int[] p;
    private int[][] grid;
    private int m;
    private int n;

    public boolean hasValidPath(int[][] grid) {
        this.grid = grid;
        m = grid.length;
        n = grid[0].length;
        p = new int[m * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int e = grid[i][j];
                if (e == 1) {
                    left(i, j);
                    right(i, j);
                } else if (e == 2) {
                    up(i, j);
                    down(i, j);
                } else if (e == 3) {
                    left(i, j);
                    down(i, j);
                } else if (e == 4) {
                    right(i, j);
                    down(i, j);
                } else if (e == 5) {
                    left(i, j);
                    up(i, j);
                } else {
                    right(i, j);
                    up(i, j);
                }
            }
        }
        return find(0) == find(m * n - 1);
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    private void left(int i, int j) {
        if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
            p[find(i * n + j)] = find(i * n + j - 1);
        }
    }

    private void right(int i, int j) {
        if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
            p[find(i * n + j)] = find(i * n + j + 1);
        }
    }

    private void up(int i, int j) {
        if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
            p[find(i * n + j)] = find((i - 1) * n + j);
        }
    }

    private void down(int i, int j) {
        if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
            p[find(i * n + j)] = find((i + 1) * n + j);
        }
    }
}

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class Solution {
public:
    vector<int> p;

    bool hasValidPath(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        p.resize(m * n);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        auto left = [&](int i, int j) {
            if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
                p[find(i * n + j)] = find(i * n + j - 1);
            }
        };
        auto right = [&](int i, int j) {
            if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
                p[find(i * n + j)] = find(i * n + j + 1);
            }
        };
        auto up = [&](int i, int j) {
            if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
                p[find(i * n + j)] = find((i - 1) * n + j);
            }
        };
        auto down = [&](int i, int j) {
            if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
                p[find(i * n + j)] = find((i + 1) * n + j);
            }
        };
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int e = grid[i][j];
                if (e == 1) {
                    left(i, j);
                    right(i, j);
                } else if (e == 2) {
                    up(i, j);
                    down(i, j);
                } else if (e == 3) {
                    left(i, j);
                    down(i, j);
                } else if (e == 4) {
                    right(i, j);
                    down(i, j);
                } else if (e == 5) {
                    left(i, j);
                    up(i, j);
                } else {
                    right(i, j);
                    up(i, j);
                }
            }
        }
        return find(0) == find(m * n - 1);
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

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func hasValidPath(grid [][]int) bool {
    m, n := len(grid), len(grid[0])
    p := make([]int, m*n)
    for i := range p {
        p[i] = i
    }
    var find func(x int) int
    find = func(x int) int {
        if p[x] != x {
            p[x] = find(p[x])
        }
        return p[x]
    }
    left := func(i, j int) {
        if j > 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) {
            p[find(i*n+j)] = find(i*n + j - 1)
        }
    }
    right := func(i, j int) {
        if j < n-1 && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) {
            p[find(i*n+j)] = find(i*n + j + 1)
        }
    }
    up := func(i, j int) {
        if i > 0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4) {
            p[find(i*n+j)] = find((i-1)*n + j)
        }
    }
    down := func(i, j int) {
        if i < m-1 && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) {
            p[find(i*n+j)] = find((i+1)*n + j)
        }
    }
    for i, row := range grid {
        for j, e := range row {
            if e == 1 {
                left(i, j)
                right(i, j)
            } else if e == 2 {
                up(i, j)
                down(i, j)
            } else if e == 3 {
                left(i, j)
                down(i, j)
            } else if e == 4 {
                right(i, j)
                down(i, j)
            } else if e == 5 {
                left(i, j)
                up(i, j)
            } else {
                right(i, j)
                up(i, j)
            }
        }
    }
    return find(0) == find(m*n-1)
}

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