Added python solution for leetcode0980

Author: soham0-0

algorithms/python/uniquePaths/uniquePathsIII.py

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+# 980. Unique Paths III
+# 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
+
+class Solution:
+    ans = 0
+    def findPathNum(self, i, j, grid: List[List[int]], curLen, pLen)->None:
+        if(grid[i][j]==2):
+            if(pLen-1==curLen):
+                self.ans+=1
+            return
+        elif (grid[i][j]==-1):
+            return
+        curLen+=1
+        grid[i][j]=-1
+        if(i-1>=0):
+            self.findPathNum(i-1, j, grid, curLen, pLen)
+        if(j-1>=0):
+            self.findPathNum(i, j-1, grid, curLen, pLen)
+        if(i+1<len(grid)):
+            self.findPathNum(i+1, j, grid, curLen, pLen)
+        if(j+1<len(grid[0])):
+            self.findPathNum(i, j+1, grid, curLen, pLen)
+        grid[i][j]=0
+        
+            
+    def uniquePathsIII(self, grid: List[List[int]]) -> int:
+        pathLen = 0
+        start = (0, 0)
+        for i in range(len(grid)):
+            for j in range(len(grid[0])):
+                if(grid[i][j]!=-1):
+                    pathLen+=1
+                    if(grid[i][j]==1):
+                        start = (i, j)
+        self.findPathNum(start[0], start[1], grid, 0, pathLen)
+        return self.ans
+