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980. Unique Paths III

Difficulty Topics

Description

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

unique-paths-iii.py
class Solution:
    def uniquePathsIII(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        fullMask = (1 << (rows * cols)) - 1
        sx, sy = (-1, -1)
        mask = 0

        for x in range(rows):
            for y in range(cols):
                if grid[x][y] == 1:
                    sx, sy = x, y
                    mask ^= (1 << (sx * cols + sy))

                if grid[x][y] == -1:
                    mask ^= (1 << (x * cols + y))        

        @cache
        def go(x, y, mask):
            if grid[x][y] == 2:
                return 1 if mask == fullMask else 0

            res = 0
            for dx, dy in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
                if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] != -1 and mask & (1 << (dx * cols + dy)) == 0:
                    res += go(dx, dy, mask ^ (1 << (dx * cols + dy)))

            return res

        return go(sx, sy, mask)