The challenge
On a 2-dimensional grid, there are 4 types of squares:
1represents the starting square. There is exactly one starting square.2represents the ending square. There is exactly one ending square.0represents empty squares we can walk over.-1represents 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 <= 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][c] == 1){
sx = r;
sy = c;
// otherwise count the zeros
} else if(grid[r][c] == 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;
}
}