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);
}
};
Java
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;
}
}