How to Solve Unique Paths III in Java

The challenge

On a 2-dimensional grid, there are 4 types of squares:

• 1 represents the starting square.  There is exactly one starting square.
• 2 represents the ending square.  There is exactly one ending square.
• 0 represents empty squares we can walk over.
• -1 represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Example 1:

Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: [[0,1],[2,0]]
Output: 0
Explanation: 
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

Note:

1. 1 <= grid.length * grid[0].length <= 20

The solution in Java

This solution uses Depth First Search to come to an answer.

class Solution {
    public int uniquePathsIII(int[][] grid) {
        
        /* 
           1 = start
           2 = end
           0 = empty
          -1 = obstacle
        [
            [1,0,0,0],
            [0,0,0,0],
            [0,0,2,-1]
        ]
        */
        
        // start `x`
        int sx = 0;
        // start `y`
        int sy = 0;
        // start `zero`
        int zero = 0;
        
        // loop through grid columns
        for(int r = 0; r<grid.length; r++) {
            // loop through grid's first row
            for(int c = 0; c<grid[0].length; c++) {
                // find where the starting point is
                if(grid[r] == 1){
                    sx = r;
                    sy = c;
                    
                // otherwise count the zeros
                } else if(grid[r] == 0) zero++;
            }
        }
        
        // return from Depth First Search
        return dfs(grid, sx, sy, zero);
    }
    
    // dfs = Depth First Search
    private int dfs(int[][] grid, int x, int y, int zero) {
        // if out of bounds, return
        if( x < 0 || y < 0 || x >= grid.length || y >= grid[0].length || grid[x][y] == -1){
            return 0;
        }
        
        // if end, return
        if(grid[x][y] == 2) {
            return zero == -1 ? 1 : 0;
        }
        // set current points to obstacle / something we can't go over again
        grid[x][y] = -1;
        // decrement the amount of zeros
        zero--;
        
        // recurse in all directions
        int totalPaths = dfs(grid, x+1, y,   zero) +
                         dfs(grid, x,   y+1, zero) +
                         dfs(grid, x-1, y,   zero) +
                         dfs(grid, x,   y-1, zero);
        
        // set to a path
        grid[x][y] = 0;
        // increment the amount of zeros
        zero++;
        
        // return the amount of paths
        return totalPaths;
    }
}