74 lines
2.4 KiB
Python
74 lines
2.4 KiB
Python
# 980. Unique Paths III
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# On a 2-dimensional grid, there are 4 types of squares:
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# 1 represents the starting square. There is exactly one starting square.
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# 2 represents the ending square. There is exactly one ending square.
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# 0 represents empty squares we can walk over.
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# -1 represents obstacles that we cannot walk over.
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# Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.
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# Example 1:
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# Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
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# Output: 2
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# Explanation: We have the following two paths:
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# 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
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# 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
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# Example 2:
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# Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
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# Output: 4
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# Explanation: We have the following four paths:
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# 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)
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# 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)
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# 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)
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# 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)
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# Example 3:
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# Input: [[0,1],[2,0]]
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# Output: 0
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# Explanation:
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# There is no path that walks over every empty square exactly once.
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# Note that the starting and ending square can be anywhere in the grid.
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# Note:
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# 1 <= grid.length * grid[0].length <= 20
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class Solution:
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ans = 0
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def findPathNum(self, i, j, grid: List[List[int]], curLen, pLen)->None:
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if(grid[i][j]==2):
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if(pLen-1==curLen):
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self.ans+=1
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return
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elif (grid[i][j]==-1):
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return
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curLen+=1
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grid[i][j]=-1
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if(i-1>=0):
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self.findPathNum(i-1, j, grid, curLen, pLen)
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if(j-1>=0):
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self.findPathNum(i, j-1, grid, curLen, pLen)
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if(i+1<len(grid)):
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self.findPathNum(i+1, j, grid, curLen, pLen)
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if(j+1<len(grid[0])):
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self.findPathNum(i, j+1, grid, curLen, pLen)
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grid[i][j]=0
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def uniquePathsIII(self, grid: List[List[int]]) -> int:
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pathLen = 0
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start = (0, 0)
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for i in range(len(grid)):
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for j in range(len(grid[0])):
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if(grid[i][j]!=-1):
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pathLen+=1
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if(grid[i][j]==1):
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start = (i, j)
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self.findPathNum(start[0], start[1], grid, 0, pathLen)
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return self.ans
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