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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.lengthn == grid[i].length1 <= m, n <= 3001 <= 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) }