# 63. Unique Paths II

## 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.

## Solutions

Solution 1: Dynamic Programming

We define $dp[i][j]$ to represent the number of paths to reach the grid $(i,j)$.

First, initialize all values in the first column and first row of $dp$, then traverse other rows and columns, there are two cases:

• If $obstacleGrid[i][j] = 1$, it means the number of paths is $0$, so $dp[i][j] = 0$;
• If $obstacleGrid[i][j] = 0$, then $dp[i][j] = dp[i - 1][j] + dp[i][j - 1]$.

Finally, return $dp[m - 1][n - 1]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

• class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length, n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
dp[i][0] = 1;
}
for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
dp[0][j] = 1;
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (obstacleGrid[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
}

• class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size(), n = obstacleGrid[0].size();
vector<vector<int>> dp(m, vector<int>(n));
for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
dp[i][0] = 1;
}
for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
dp[0][j] = 1;
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (obstacleGrid[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};

• class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
if not obstacleGrid:
return 0

m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * (n+1) for _ in range(m+1)]

# dp[1][1] = dp[0][1] + dp[1][0]
# so, set either dp[0][1]=1, or set dp[1][0]=1
dp[0][1] = 1

for i in range(1, m+1):
for j in range(1, n+1):
if obstacleGrid[i-1][j-1] == 0:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
# else, ==1, obstacle, skip and leave as 0

return dp[m][n]

############

class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * n for _ in range(m)]
for i in range(m):
if obstacleGrid[i][0] == 1:
break
dp[i][0] = 1
for j in range(n):
if obstacleGrid[0][j] == 1:
break
dp[0][j] = 1
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j] == 0:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
return dp[-1][-1]


• func uniquePathsWithObstacles(obstacleGrid [][]int) int {
m, n := len(obstacleGrid), len(obstacleGrid[0])
dp := make([][]int, m)
for i := 0; i < m; i++ {
dp[i] = make([]int, n)
}
for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
dp[i][0] = 1
}
for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {
dp[0][j] = 1
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
if obstacleGrid[i][j] == 0 {
dp[i][j] = dp[i-1][j] + dp[i][j-1]
}
}
}
return dp[m-1][n-1]
}

• function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
const m = obstacleGrid.length;
const n = obstacleGrid[0].length;
const dp = Array.from({ length: m }, () => new Array(n).fill(0));
for (let i = 0; i < m; i++) {
if (obstacleGrid[i][0] === 1) {
break;
}
dp[i][0] = 1;
}
for (let i = 0; i < n; i++) {
if (obstacleGrid[0][i] === 1) {
break;
}
dp[0][i] = 1;
}
for (let i = 1; i < m; i++) {
for (let j = 1; j < n; j++) {
if (obstacleGrid[i][j] === 1) {
continue;
}
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}


• impl Solution {
pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
let m = obstacle_grid.len();
let n = obstacle_grid[0].len();
if obstacle_grid[0][0] == 1 || obstacle_grid[m - 1][n - 1] == 1 {
return 0;
}
let mut dp = vec![vec![0; n]; m];
for i in 0..n {
if obstacle_grid[0][i] == 1 {
break;
}
dp[0][i] = 1;
}
for i in 0..m {
if obstacle_grid[i][0] == 1 {
break;
}
dp[i][0] = 1;
}
for i in 1..m {
for j in 1..n {
if obstacle_grid[i][j] == 1 {
continue;
}
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
dp[m - 1][n - 1]
}
}