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: Memoization Search

We design a function \(\textit{dfs}(i, j)\) to represent the number of paths from the grid \((i, j)\) to the grid \((m - 1, n - 1)\). Here, \(m\) and \(n\) are the number of rows and columns of the grid, respectively.

The execution process of the function \(\textit{dfs}(i, j)\) is as follows:

If \(i \ge m\) or \(j \ge n\), or \(\textit{obstacleGrid}[i][j] = 1\), the number of paths is \(0\);
If \(i = m - 1\) and \(j = n - 1\), the number of paths is \(1\);
Otherwise, the number of paths is \(\textit{dfs}(i + 1, j) + \textit{dfs}(i, j + 1)\).

To avoid redundant calculations, we can use memoization.

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.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= m or j >= n or obstacleGrid[i][j]:
                return 0
            if i == m - 1 and j == n - 1:
                return 1
            return dfs(i + 1, j) + dfs(i, j + 1)

        m, n = len(obstacleGrid), len(obstacleGrid[0])
        return dfs(0, 0)

class Solution {
    private Integer[][] f;
    private int[][] obstacleGrid;
    private int m;
    private int n;

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        m = obstacleGrid.length;
        n = obstacleGrid[0].length;
        this.obstacleGrid = obstacleGrid;
        f = new Integer[m][n];
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= m || j >= n || obstacleGrid[i][j] == 1) {
            return 0;
        }
        if (i == m - 1 && j == n - 1) {
            return 1;
        }
        if (f[i][j] == null) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    }
}

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

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m, n := len(obstacleGrid), len(obstacleGrid[0])
    f := make([][]int, m)
    for i := range f {
        f[i] = make([]int, n)
        for j := range f[i] {
            f[i][j] = -1
        }
    }
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if i >= m || j >= n || obstacleGrid[i][j] == 1 {
            return 0
        }
        if i == m-1 && j == n-1 {
            return 1
        }
        if f[i][j] == -1 {
            f[i][j] = dfs(i+1, j) + dfs(i, j+1)
        }
        return f[i][j]
    }
    return dfs(0, 0)
}

function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
            return 0;
        }
        if (i === m - 1 && j === n - 1) {
            return 1;
        }
        if (f[i][j] === -1) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    };
    return dfs(0, 0);
}

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();
        let mut f = vec![vec![-1; n]; m];
        Self::dfs(0, 0, &obstacle_grid, &mut f)
    }

    fn dfs(i: usize, j: usize, obstacle_grid: &Vec<Vec<i32>>, f: &mut Vec<Vec<i32>>) -> i32 {
        let m = obstacle_grid.len();
        let n = obstacle_grid[0].len();
        if i >= m || j >= n || obstacle_grid[i][j] == 1 {
            return 0;
        }
        if i == m - 1 && j == n - 1 {
            return 1;
        }
        if f[i][j] != -1 {
            return f[i][j];
        }
        let down = Self::dfs(i + 1, j, obstacle_grid, f);
        let right = Self::dfs(i, j + 1, obstacle_grid, f);
        f[i][j] = down + right;
        f[i][j]
    }
}

/**
 * @param {number[][]} obstacleGrid
 * @return {number}
 */
var uniquePathsWithObstacles = function (obstacleGrid) {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    const f = Array.from({ length: m }, () => Array(n).fill(-1));
    const dfs = (i, j) => {
        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
            return 0;
        }
        if (i === m - 1 && j === n - 1) {
            return 1;
        }
        if (f[i][j] === -1) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    };
    return dfs(0, 0);
};

Solution 2: Dynamic Programming

We can use a dynamic programming approach by defining a 2D array \(f\), where \(f[i][j]\) represents the number of paths from the grid \((0,0)\) to the grid \((i,j)\).

We first initialize all values in the first column and the first row of \(f\), then traverse the other rows and columns with two cases:

If \(\textit{obstacleGrid}[i][j] = 1\), it means the number of paths is \(0\), so \(f[i][j] = 0\);
If \(\textit{obstacleGrid}[i][j] = 0\), then \(f[i][j] = f[i - 1][j] + f[i][j - 1]\).

Finally, return \(f[m - 1][n - 1]\).