On a 2-dimensional grid, there are 4 types of squares:
1 represents the starting square. There is exactly one starting square.
2 represents the ending square. There is exactly one ending square.
0 represents empty squares we can walk over.
-1 represents obstacles that we cannot walk over.
Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.
Example 1:
Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths:
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
Example 2:
Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths:
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
Example 3:
Input: [[0,1],[2,0]]
Output: 0
Explanation:
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.
Note:
1 <= grid.length * grid[0].length <= 20
// Recursive DFS
void dfs(vector<vector<int> >& grid, vector<vector<bool> >& visited, int x, int y, int end_x, int end_y, int& zeros, int& res) {
if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] < 0 || visited[x][y]) return;
if (x == end_x && y == end_y) {
if (zeros == 0) ++res;
return;
}
visited[x][y] = true;
--zeros;
dfs(grid, visited, x + 1, y, end_x, end_y, zeros, res);
dfs(grid, visited, x - 1, y, end_x, end_y, zeros, res);
dfs(grid, visited, x, y + 1, end_x, end_y, zeros, res);
dfs(grid, visited, x, y - 1, end_x, end_y, zeros, res);
++zeros;
visited[x][y] = false;
}
int uniquePathsIII(vector<vector<int>>& grid) { // time: O(4^(mn)); space: O(mn)
int m = grid.size(), n = grid[0].size();
int res = 0, zeros = 0, start_x = -1, start_y = -1, end_x = -1, end_y = -1;
vector<vector<bool> > visited(m, vector<bool> (n, false));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) ++zeros;
else if (grid[i][j] == 1) {
++zeros;
start_x = i;
start_y = j;
} else if (grid[i][j] == 2) {
end_x = i;
end_y = j;
}
}
}
dfs(grid, visited, start_x, start_y, end_x, end_y, zeros, res);
return res;
}
// Brute Force Recursion
bool check(vector<vector<int>>& grid, int x, int y) {
int m = grid.size(), n = grid[0].size();
return 0 <= x && x < m && 0 <= y && y < n && grid[x][y] >= 0;
}
void dfs(vector<vector<int>>& grid, int x, int y, int end_x, int end_y, int& zeros, int& res) {
if (!check(grid, x, y)) return;
if (x == end_x && y == end_y) {
if (zeros == 0) ++res;
return;
}
grid[x][y] = -1;
--zeros;
dfs(grid, x + 1, y, end_x, end_y, zeros, res);
dfs(grid, x - 1, y, end_x, end_y, zeros, res);
dfs(grid, x, y + 1, end_x, end_y, zeros, res);
dfs(grid, x, y - 1, end_x, end_y, zeros, res);
++zeros;
grid[x][y] = 0;
}
int uniquePathsIII(vector<vector<int>>& grid) { // time: O(4^(mn)); space: O(mn)
int res = 0, zeros = 1, start_x = -1, start_y = -1, end_x = -1, end_y = -1;
int m = grid.size(), n = grid[0].size();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) ++zeros;
else if (grid[i][j] == 1) {
start_x = i;
start_y = j;
} else if (grid[i][j] == 2) {
end_x = i;
end_y = j;
}
}
}
dfs(grid, start_x, start_y, end_x, end_y, zeros, res);
return res;
}