# 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.

```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```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
}
}
```Code language: Java (java)```
Tags:
Subscribe
Notify of