Question

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

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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

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)

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);
    }
  }
}

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];
  }
};

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)
}