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Formatted question description: https://leetcode.ca/all/1391.html

1391. Check if There is a Valid Path in a Grid (Medium)

Given a m x n grid. Each cell of the 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 which 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).

Example 4:

Input: grid = [[1,1,1,1,1,1,3]]
Output: true

Example 5:

Input: grid = [[2],[2],[2],[2],[2],[2],[6]]
Output: true

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • 1 <= grid[i][j] <= 6

Related Topics: Depth-first Search, Breadth-first Search

Solution 1. Union Find

// OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
// Time: O(MN)
// Space: O(MN)
class UnionFind {
private:
    vector<int> id, rank;
    int find (int i) {
        if (id[i] == i) return i;
        return id[i] = find(id[i]);
    }
public:
    UnionFind(int n) : id(n), rank(n, 0) {
        for (int i = 0; i < n; ++i) id[i] = i;
    }
    void connect(int i, int j) {
        int p = find(i), q = find(j);
        if (p == q) return;
        if (rank[p] > rank[q]) id[p] = q;
        else {
            id[q] = p;
            if (rank[p] == rank[q]) rank[p]++;
        }
    }
    bool connected(int i, int j) { return find(i) == find(j); }
};
class Solution {
    int M, N;
    int h(int x, int y) { return x * N + y; }
    const int dirs[4][2] = { {0,-1},{0,1},{-1,0},{1,0} };
    const int neighbor[6][2] = { {0,1}, {2,3}, {0,3}, {1,3}, {0,2}, {1,2} };
public:
    bool hasValidPath(vector<vector<int>>& A) {
        M = A.size(), N = A[0].size();
        UnionFind uf(M * N);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                for (int n : neighbor[A[i][j] - 1]) {
                    int x = i + dirs[n][0], y = j + dirs[n][1];
                    if (x < 0 || x >= M || y < 0 || y >= N) continue;
                    int r = n ^ 1;
                    auto &rn = neighbor[A[x][y] - 1];
                    if (rn[0] != r && rn[1] != r) continue;
                    uf.connect(h(x, y), h(i, j));
                }
            }
        }
        return uf.connected(h(0,0), h(M-1,N-1));
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
// Time: O(MN)
// Space: O(MN)
// Ref: https://www.youtube.com/watch?v=SpMez87v0O8
class Solution {
    int M, N;
    vector<vector<int>> A;
    vector<vector<bool>> vis;
    int h(int x, int y) { return x * N + y; }
    int dx[4] = {1,-1,0,0}, dy[4] = {0,0,1,-1}, t[6] = {4|8, 1|2, 8|1, 4|1, 8|2, 4|2};
    bool dfs(int i, int j) {
        if (i == M - 1 && j == N - 1) return 1;
        vis[i][j] = 1;
        for (int k = 0; k < 4; ++k) {
            if (t[A[i][j] - 1] >> k & 1 ^ 1) continue; // If A[i][j] can't extend to this direction, skip
            int x = i + dx[k], y = j + dy[k];
            if (x < 0 || x >= M || y < 0 || y >= N || vis[x][y]) continue;
            int rk = k ^ 1;
            if (t[A[x][y] - 1] >> rk & 1 ^ 1) continue; // If A[x][y] can't extend back, skip
            if (dfs(x, y)) return 1;
        }
        return 0;
    }
public:
    bool hasValidPath(vector<vector<int>>& A) {
        M = A.size(), N = A[0].size();
        this->A = A;
        vis.assign(M, vector<bool>(N));
        return dfs(0, 0);
    }
};
  • class Solution {
        public boolean hasValidPath(int[][] grid) {
            int rows = grid.length, columns = grid[0].length;
            boolean[][] visited = new boolean[rows][columns];
            visited[0][0] = true;
            Queue<int[]> queue = new LinkedList<int[]>();
            queue.offer(new int[]{0, 0});
            while (!queue.isEmpty()) {
                int[] cell = queue.poll();
                if (cell[0] == rows - 1 && cell[1] == columns - 1)
                    return true;
                List<int[]> reachableCells = reachableCells(grid[cell[0]][cell[1]], grid, cell, rows, columns);
                for (int[] reachableCell : reachableCells) {
                    int row = reachableCell[0], column = reachableCell[1];
                    if (!visited[row][column]) {
                        visited[row][column] = true;
                        queue.offer(new int[]{row, column});
                    }
                }
            }
            return false;
        }
    
        public List<int[]> reachableCells(int street, int[][] grid, int[] cell, int rows, int columns) {
            List<int[]> reachableCells = new ArrayList<int[]>();
            int row = cell[0], column = cell[1];
            if (street == 1) {
                int newColumn1 = column - 1, newColumn2 = column + 1;
                if (newColumn1 >= 0) {
                    int newStreet = grid[row][newColumn1];
                    if (newStreet == 1 || newStreet == 4 || newStreet == 6)
                        reachableCells.add(new int[]{row, newColumn1});
                }
                if (newColumn2 < columns) {
                    int newStreet = grid[row][newColumn2];
                    if (newStreet == 1 || newStreet == 3 || newStreet == 5)
                        reachableCells.add(new int[]{row, newColumn2});
                }
            } else if (street == 2) {
                int newRow1 = row - 1, newRow2 = row + 1;
                if (newRow1 >= 0) {
                    int newStreet = grid[newRow1][column];
                    if (newStreet == 2 || newStreet == 3 || newStreet == 4)
                        reachableCells.add(new int[]{newRow1, column});
                }
                if (newRow2 < rows) {
                    int newStreet = grid[newRow2][column];
                    if (newStreet == 2 || newStreet == 5 || newStreet == 6)
                        reachableCells.add(new int[]{newRow2, column});
                }
            } else if (street == 3) {
                int newRow = row + 1, newColumn = column - 1;
                if (newRow < rows) {
                    int newStreet = grid[newRow][column];
                    if (newStreet == 2 || newStreet == 5 || newStreet == 6)
                        reachableCells.add(new int[]{newRow, column});
                }
                if (newColumn >= 0) {
                    int newStreet = grid[row][newColumn];
                    if (newStreet == 1 || newStreet == 4 || newStreet == 6)
                        reachableCells.add(new int[]{row, newColumn});
                }
            } else if (street == 4) {
                int newRow = row + 1, newColumn = column + 1;
                if (newRow < rows) {
                    int newStreet = grid[newRow][column];
                    if (newStreet == 2 || newStreet == 5 || newStreet == 6)
                        reachableCells.add(new int[]{newRow, column});
                }
                if (newColumn < columns) {
                    int newStreet = grid[row][newColumn];
                    if (newStreet == 1 || newStreet == 3 || newStreet == 5)
                        reachableCells.add(new int[]{row, newColumn});
                }
            } else if (street == 5) {
                int newRow = row - 1, newColumn = column - 1;
                if (newRow >= 0) {
                    int newStreet = grid[newRow][column];
                    if (newStreet == 2 || newStreet == 3 || newStreet == 4)
                        reachableCells.add(new int[]{newRow, column});
                }
                if (newColumn >= 0) {
                    int newStreet = grid[row][newColumn];
                    if (newStreet == 1 || newStreet == 4 || newStreet == 6)
                        reachableCells.add(new int[]{row, newColumn});
                }
            } else if (street == 6) {
                int newRow = row - 1, newColumn = column + 1;
                if (newRow >= 0) {
                    int newStreet = grid[newRow][column];
                    if (newStreet == 2 || newStreet == 3 || newStreet == 4)
                        reachableCells.add(new int[]{newRow, column});
                }
                if (newColumn < columns) {
                    int newStreet = grid[row][newColumn];
                    if (newStreet == 1 || newStreet == 3 || newStreet == 5)
                        reachableCells.add(new int[]{row, newColumn});
                }
            }
            return reachableCells;
        }
    }
    
    ############
    
    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);
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/check-if-there-is-a-valid-path-in-a-grid/
    // Time: O(MN)
    // Space: O(MN)
    class UnionFind {
    private:
        vector<int> id, rank;
        int find (int i) {
            if (id[i] == i) return i;
            return id[i] = find(id[i]);
        }
    public:
        UnionFind(int n) : id(n), rank(n, 0) {
            for (int i = 0; i < n; ++i) id[i] = i;
        }
        void connect(int i, int j) {
            int p = find(i), q = find(j);
            if (p == q) return;
            if (rank[p] > rank[q]) id[p] = q;
            else {
                id[q] = p;
                if (rank[p] == rank[q]) rank[p]++;
            }
        }
        bool connected(int i, int j) { return find(i) == find(j); }
    };
    class Solution {
        int M, N;
        int h(int x, int y) { return x * N + y; }
        const int dirs[4][2] = { {0,-1},{0,1},{-1,0},{1,0} };
        const int neighbor[6][2] = { {0,1}, {2,3}, {0,3}, {1,3}, {0,2}, {1,2} };
    public:
        bool hasValidPath(vector<vector<int>>& A) {
            M = A.size(), N = A[0].size();
            UnionFind uf(M * N);
            for (int i = 0; i < M; ++i) {
                for (int j = 0; j < N; ++j) {
                    for (int n : neighbor[A[i][j] - 1]) {
                        int x = i + dirs[n][0], y = j + dirs[n][1];
                        if (x < 0 || x >= M || y < 0 || y >= N) continue;
                        int r = n ^ 1;
                        auto &rn = neighbor[A[x][y] - 1];
                        if (rn[0] != r && rn[1] != r) continue;
                        uf.connect(h(x, y), h(i, j));
                    }
                }
            }
            return uf.connected(h(0,0), h(M-1,N-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)
    
    
    
  • 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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