63. Unique Paths II (Medium)

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

Solution: DP

Denote the number of paths to arrive at point (i, j) to be P[i][j], the state equation is P[i][j] = P[i - 1][j] + P[i][j - 1] if obstacleGrid[i][j] != 1 and 0 otherwise.

version 1: 3ms Time Complexity: $$O(mn)$$ Space Complexity: $$O(mn)$$

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) return 0;
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<vector<int>> dp(m,vector<int>(n,0));

        for (int i = 0; i < n; ++i) {
            if (obstacleGrid[0][i] == 1) break;
            dp[0][i] = 1;
        }

        for (int i = 0; i < m; ++i) {
            if (obstacleGrid[i][0] == 1) break;
            dp[i][0] = 1;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                dp[i][j] = obstacleGrid[i][j] == 1 ? 0:dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};

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

version 2: 3ms Time Complexity: $$O(m*n)$$ Space Complexity: $$O(n)$$

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