LeetCode: Unique Paths II

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.

---

The same idea as Unique Paths. We can just add one more check on the current grid spot if it is an obstacle. If it is, set grid[i][j] to 0.

C++ Code:

/*
 * func: unique_path_with_obstacle
 * goal: to get the total number of paths while obstacles exist
 * @param obstacleGrid: the grid with obstacle in
 * return: total paths
 */
/*
 * The same way as uniquePath, just add one more condition to check if 
 * current spot is an obstacle one
 * complexity: time O(m*n), space O(m*n)
 */
int unique_path_with_obstacle(const vector<vector<int> > &obstacleGrid){
    if(obstacleGrid.size() == 0){
        return 0;
    }
    int m = obstacleGrid.size();
    int n = obstacleGrid[0].size();
    int **grid =  new int* [m];
    for(int i=0; i<m; i++){
        grid[i] = new int[n];
    }
    grid[0][0]=1;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(i == 0 && j > 0){
                grid[i][j] = grid[i][j-1];
            }else if(i > 0 && j == 0){
                grid[i][j] = grid[i-1][j];
            }else if(i > 0 && j > 0){
                grid[i][j] = grid[i-1][j] + grid[i][j-1];
            }
            if(obstacleGrid[i][j] == 1){
                grid[i][j] = 0;
            }
        }
    }
    int result = grid[m-1][n-1];
    for(int i=0; i<m; i++){
        delete[] grid[i];
    }
    delete [] grid;
    return result;

}

Python Code:

# func: find the number of total unique paths to the target spot while obstacles exist
# @param obstacleGrid: the number of rows
# @return: total numbers
# complexity: time O(m*n), space O(m*n)
def unique_paths(obstacleGrid):
    if not obstacleGrid:
        return 0

    m = len(obstacleGrid)
    n = len(obstacleGrid[0])
    grid = [[1] * n for _ in xrange(m)]
    for i in xrange(m):
        for j in xrange(n):
            if i > 0 and j == 0:
                grid[i][j] = grid[i-1][j]
            elif i == 0 and j > 0:
                grid[i][j] = grid[i][j-1]
            elif i > 0 and j > 0:
                grid[i][j] = grid[i-1][j] + grid[i][j-1]
            if obstacleGrid[i][j] == 1:
                grid[i][j] = 0

    return grid[m-1][n-1]