Question

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.

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:
  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]

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];
  }
};

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]
  }
}