980. Unique Paths III

Problem Statement

Hard

You are given an m x n integer array grid where grid[i][j] could be:

1 representing the starting square. There is exactly one starting square.
2 representing the ending square. There is exactly one ending square.
0 representing empty squares we can walk over.
-1 representing 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: grid = [[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: grid = [[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: grid = [[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.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 20
1 <= m * n <= 20
-1 <= grid[i][j] <= 2
There is exactly one starting cell and one ending cell.

C++

Source file

class Solution {
public:
    int res = 0, start_i, start_j, box = 1;
    
    bool isInvalid(vector<vector<int>>& grid, int i, int j){
        return (i<0 || j<0 || i>=grid.size() || j>=grid[i].size());
    }
    
    void dfs(vector<vector<int>> grid, int i, int j, int ct=0){
        if(isInvalid(grid, i, j) || grid[i][j]==-1)
            return;
        if(grid[i][j]==2){
            if(ct==box)
                res++;
            return;
        }
        grid[i][j] = -1;
        dfs(grid, i, j+1, ct+1);
        dfs(grid, i, j-1, ct+1);
        dfs(grid, i+1, j, ct+1);
        dfs(grid, i-1, j, ct+1);
        grid[i][j] = 0;
    }
    
    int uniquePathsIII(vector<vector<int>>& grid) {
        for(int i=0; i<grid.size(); i++){
            for(int j=0; j<grid[i].size(); j++){
                if(grid[i][j]==1)
                    start_i = i, start_j = j;
                else if(!grid[i][j]) box++;
            }
        }
        dfs(grid, start_i, start_j);
        return res;
    }
};