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:

 1 // 1WA, 1AC

 2 class Solution {

 3 public:

 4     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         int m, n;

 8         

 9         m = obstacleGrid.size();

10         if(m <= 0){

11             return 0;

12         }

13         

14         n = obstacleGrid[0].size();

15         

16         int **arr = nullptr;

17         int i, j;

18         

19         arr = new int*[m];

20         for(i = 0; i < m; ++i){

21             arr[i] = new int[n];

22         }

23         

24         arr[0][0] = obstacleGrid[0][0] ? 0 : 1;

25         for(i = 1; i < m; ++i){

26             // 1WA here, arr[i][0] = obstacleGrid[i][0] ? 0 : obstacleGrid[i - 1][0];

27             // arr[i][0] = obstacleGrid[i][0] ? 0 : arr[i - 1][0];

28             arr[i][0] = obstacleGrid[i][0] ? 0 : arr[i - 1][0];

29         }

30         for(i = 1; i < n; ++i){

31             arr[0][i] = obstacleGrid[0][i] ? 0 : arr[0][i - 1];

32         }

33         for(i = 1; i < m; ++i){

34             for(j = 1; j < n; ++j){

35                 arr[i][j] = obstacleGrid[i][j] ? 0 : arr[i - 1][j] + arr[i][j - 1];

36             }

37         }

38         

39         int res = arr[m - 1][n - 1];

40         for(i = 0; i < m; ++i){

41             delete[] arr[i];

42         }

43         delete[] arr;

44         

45         return res;

46     }

47 };

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:

 1 // 1AC, very nice

 2 class Solution {

 3 public:

 4     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         int m, n;

 8         

 9         m = obstacleGrid.size();

10         if(m <= 0){

11             return 0;

12         }

13         

14         n = obstacleGrid[0].size();

15         

16         int **arr = nullptr;

17         int i, j;

18         int flag = 0;

19         

20         arr = new int*[2];

21         for(i = 0; i < 2; ++i){

22             arr[i] = new int[n];

23         }

24         

25         flag = 0;

26         arr[flag][0] = obstacleGrid[0][0] ? 0 : 1;

27         for(i = 1; i < n; ++i){

28             arr[flag][i] = obstacleGrid[0][i] ? 0 : arr[flag][i - 1];

29         }

30         for(i = 1; i < m; ++i){

31             flag = !flag;

32             arr[flag][0] = obstacleGrid[i][0] ? 0 : arr[!flag][0];

33             for(j = 1; j < n; ++j){

34                 arr[flag][j] = obstacleGrid[i][j] ? 0 : arr[!flag][j] + arr[flag][j - 1];

35             }

36         }

37         

38         j = arr[flag][n - 1];

39         for(i = 0; i < 2; ++i){

40             delete[] arr[i];

41         }

42         delete[] arr;

43         

44         return j;

45     }

46 };