You are given an m x n
integer array grid
where grid[i][j]
could be:
-
1
representing the starting square. There is exactly one starting square. -
2
representing the ending square. There is exactly one ending square. -
0
representing empty squares we can walk over. -
-1
representing 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: grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths:
- (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(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: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths:
- (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
- (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
- (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
- (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: grid = [[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.
Constraints:
-
m == grid.length
-
n == grid[i].length
-
1 <= m, n <= 20
-
1 <= m * n <= 20
-
-1 <= grid[i][j] <= 2
- There is exactly one starting cell and one ending cell.
SOLUTION:
class Solution:
def getNeighbors(self, start, m, n):
a, b = start
neighbors = []
if a > 0:
neighbors.append((a - 1, b))
if a < m - 1:
neighbors.append((a + 1, b))
if b > 0:
neighbors.append((a, b - 1))
if b < n - 1:
neighbors.append((a, b + 1))
return neighbors
def unipaths(self, start, grid, visited, m, n, numEmpty):
currval = grid[start[0]][start[1]]
ways = 0
if currval == -1:
return ways
if currval == 2:
if len(visited) == numEmpty + 2:
return ways + 1
return ways
neighbors = self.getNeighbors(start, m, n)
for neighbor in neighbors:
if grid[neighbor[0]][neighbor[1]] != -1 and neighbor not in visited:
ways += self.unipaths(neighbor, grid, visited.union({neighbor}), m, n, numEmpty)
return ways
def uniquePathsIII(self, grid: List[List[int]]) -> int:
m = len(grid)
n = len(grid[0])
numEmpty = 0
start = None
for i in range(m):
for j in range(n):
if grid[i][j] == 0:
numEmpty += 1
elif grid[i][j] == 1:
start = (i, j)
return self.unipaths(start, grid, {start}, m, n, numEmpty)
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