text: ##### Welcome to Subscribe On Youtube
  
  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)
  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.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)
  }
  if (newColumn2 < columns) {
  int newStreet = grid[row][newColumn2];
  if (newStreet == 1 || newStreet == 3 || newStreet == 5)
  }
  } 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)
  }
  if (newRow2 < rows) {
  int newStreet = grid[newRow2][column];
  if (newStreet == 2 || newStreet == 5 || newStreet == 6)
  }
  } 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)
  }
  if (newColumn >= 0) {
  int newStreet = grid[row][newColumn];
  if (newStreet == 1 || newStreet == 4 || newStreet == 6)
  }
  } 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)
  }
  if (newColumn < columns) {
  int newStreet = grid[row][newColumn];
  if (newStreet == 1 || newStreet == 3 || newStreet == 5)
  }
  } 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)
  }
  if (newColumn >= 0) {
  int newStreet = grid[row][newColumn];
  if (newStreet == 1 || newStreet == 4 || newStreet == 6)
  }
  } 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)
  }
  if (newColumn < columns) {
  int newStreet = grid[row][newColumn];
  if (newStreet == 1 || newStreet == 3 || newStreet == 5)
  }
  }
  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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