Unique Paths III

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:

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: grid = [[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: 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)