63-unique-paths-ii

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Description

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m-1][n-1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

 

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

 

Constraints:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j] is 0 or 1.

Solution(Python)

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        self.obstacleGrid = obstacleGrid
        return self.topdown(0, 0)

    def inplacedp(self, obstacleGrid):
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])

        # If the starting cell has an obstacle, then simply return as there would be
        # no paths to the destination.
        if obstacleGrid[0][0] == 1:
            return 0

        # Number of ways of reaching the starting cell = 1.
        obstacleGrid[0][0] = 1

        # Filling the values for the first column
        for i in range(1, m):
            obstacleGrid[i][0] = int(
                obstacleGrid[i][0] == 0 and obstacleGrid[i - 1][0] == 1
            )

        # Filling the values for the first row
        for j in range(1, n):
            obstacleGrid[0][j] = int(
                obstacleGrid[0][j] == 0 and obstacleGrid[0][j - 1] == 1
            )

        # Starting from cell(1,1) fill up the values
        # No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
        # i.e. From above and left.
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 0:
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + \
                        obstacleGrid[i][j - 1]
                else:
                    obstacleGrid[i][j] = 0

        # Return value stored in rightmost bottommost cell. That is the destination.
        return obstacleGrid[m - 1][n - 1]

    def dfs(self, obstacleGrid):
        m, n = len(obstacleGrid), len(obstacleGrid[0])

        def recur(i, j):

            if i == m - 1 and j == n - 1:
                return 1

            if i >= m or j >= n or obstacleGrid[i][j]:
                return 0

            return recur(i + 1, j) + recur(i, j + 1)

        return recur(0, 0)

    @cache
    def topdown(self, i, j):

        m, n = len(self.obstacleGrid), len(self.obstacleGrid[0])

        if i == m - 1 and j == n - 1 and not self.obstacleGrid[i][j]:
            return 1

        if i >= m or j >= n or self.obstacleGrid[i][j]:
            return 0
        return self.topdown(i + 1, j) + self.topdown(i, j + 1)