LeetCode - Unique Paths II

Unique Paths II

2013.12.21 03:19

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.

Solution1:

　　This problem is a variation from "LeetCode - Unique Paths", only difference in that there're some grids defined as impenetrable obstacles.

　　I haven't come up with a solution using combinatorics. So I just did it the old-school way, dynamic programming with recurrence relation as:

　　　　a[x][y] = b[x][y] ? 0 : a[x - 1][y] + a[x][y - 1];

　　　　where a[x][y] means the number of unique paths to (x, y), b[x][y] = 1 means (x, y) is an obstacle, while 0 for free space.

　　Time complexity is O(m * n), space complexity is O(m * n).

Accepted code:

// 1WA, 1AC
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int m, n;
        
        m = obstacleGrid.size();
        if(m <= 0){
            return 0;
        }
        
        n = obstacleGrid[0].size();
        
        int **arr = nullptr;
        int i, j;
        
        arr = new int*[m];
        for(i = 0; i < m; ++i){
            arr[i] = new int[n];
        }
        
        arr[0][0] = obstacleGrid[0][0] ? 0 : 1;
        for(i = 1; i < m; ++i){
            // 1WA here, arr[i][0] = obstacleGrid[i][0] ? 0 : obstacleGrid[i - 1][0];
            // arr[i][0] = obstacleGrid[i][0] ? 0 : arr[i - 1][0];
            arr[i][0] = obstacleGrid[i][0] ? 0 : arr[i - 1][0];
        }
        for(i = 1; i < n; ++i){
            arr[0][i] = obstacleGrid[0][i] ? 0 : arr[0][i - 1];
        }
        for(i = 1; i < m; ++i){
            for(j = 1; j < n; ++j){
                arr[i][j] = obstacleGrid[i][j] ? 0 : arr[i - 1][j] + arr[i][j - 1];
            }
        }
        
        int res = arr[m - 1][n - 1];
        for(i = 0; i < m; ++i){
            delete[] arr[i];
        }
        delete[] arr;
        
        return res;
    }
};

Solution2:

　　Space complexity can be optimized to linear, using 2 rows instead of m rows of extra space. 

　　Time complexity is O(m * n), space complexity is O(n).

Accepted code:

// 1AC, very nice
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int m, n;
        
        m = obstacleGrid.size();
        if(m <= 0){
            return 0;
        }
        
        n = obstacleGrid[0].size();
        
        int **arr = nullptr;
        int i, j;
        int flag = 0;
        
        arr = new int*[2];
        for(i = 0; i < 2; ++i){
            arr[i] = new int[n];
        }
        
        flag = 0;
        arr[flag][0] = obstacleGrid[0][0] ? 0 : 1;
        for(i = 1; i < n; ++i){
            arr[flag][i] = obstacleGrid[0][i] ? 0 : arr[flag][i - 1];
        }
        for(i = 1; i < m; ++i){
            flag = !flag;
            arr[flag][0] = obstacleGrid[i][0] ? 0 : arr[!flag][0];
            for(j = 1; j < n; ++j){
                arr[flag][j] = obstacleGrid[i][j] ? 0 : arr[!flag][j] + arr[flag][j - 1];
            }
        }
        
        j = arr[flag][n - 1];
        for(i = 0; i < 2; ++i){
            delete[] arr[i];
        }
        delete[] arr;
        
        return j;
    }
};