[Leetcode 63, Medium] Unique Paths II

Problem:

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
Analysis:

Solutions:

C++:

int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if(m == 0 && n == 0)
            return 0;
            
        if(m == 0 || n == 0) {
            return 1;
        }
            
        vector<vector<int> > num_of_paths;
        for(int row = 0; row < m; ++row) {
            vector<int> new_row;
            for(int col = 0; col < n; ++col)
                new_row.push_back(1);
                
            num_of_paths.push_back(new_row);
        }
        
        for(int i = 0; i < n; ++i) {
            if(obstacleGrid[0][i] == 1) {
                for(int j = i; j < n; ++j)
                    num_of_paths[0][j] = 0;
                    
                break;
            }
        }
        
        for(int i = 0; i < m; ++i) {
            if(obstacleGrid[i][0] == 1) {
                for(int j = i; j < m; ++j)
                    num_of_paths[j][0] = 0;
                    
                break;
            }
        }
        
        for(int i = 1; i < m; ++i) {
            for(int j = 1; j < n; ++j) {
                if(obstacleGrid[i][j] == 1)
                    num_of_paths[i][j] = 0;
                else
                    num_of_paths[i][j] = num_of_paths[i - 1][j] + num_of_paths[i][j - 1];
            }
        }
        
        return num_of_paths[m - 1][n - 1];
    }

Java:

Python: