https://leetcode.com/problems/unique-paths-ii/
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
- code
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
m = len(obstacleGrid)
n = len(obstacleGrid[0])
if obstacleGrid[0][0] == 1: return 0
obstacleGrid[0][0] = 1
for i in range(1, m):
obstacleGrid[i][0] = int(obstacleGrid[i][0] == 0 and obstacleGrid[i-1][0] == 1)
for j in range(1, n):
obstacleGrid[0][j] = int(obstacleGrid[0][j] == 0 and obstacleGrid[0][j-1] == 1)
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 obstacleGrid[-1][-1]