from __future__ import print_function
# Time:  O(m * n)
# Space: O(m + n)
#
# Follow up for "Unique Paths":
#
# Now consider if some obstacles are added to the grids. How many unique paths would there be?
#
# An obstacle and empty space is marked as 1 and 0 respectively in the grid.
#
# For example,
# There is one obstacle in the middle of a 3x3 grid as illustrated below.
#
# [
#   [0,0,0],
#   [0,1,0],
#   [0,0,0]
# ]
# The total number of unique paths is 2.
#
# Note: m and n will be at most 100.
#

class Solution:
    # @param obstacleGrid, a list of lists of integers
    # @return an integer
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        m, n = len(obstacleGrid), len(obstacleGrid[0])

        ways = [0]*n
        ways[0] = 1
        for i in xrange(m):
            if obstacleGrid[i][0] == 1:
                ways[0] = 0
            for j in xrange(n):
                if obstacleGrid[i][j] == 1:
                    ways[j] = 0
                elif j>0:
                    ways[j] += ways[j-1]
        return ways[-1]

if __name__ == "__main__":
    obstacleGrid = [
                      [0,0,0],
                      [0,1,0],
                      [0,0,0]
                   ]
    print(Solution().uniquePathsWithObstacles(obstacleGrid))