Unique Paths III | Leetcode 980 Solution

980. Unique Paths III

 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.

solution : 

class Solution { public:     int dfs(vector<vector<int>>& grid , int i , int j  , int zeros , int m , int n)     {         if(i < 0 || j < 0 || i>= m || j >= n || grid[i][j] == -1)         {             return 0;         }         if(grid[i][j] == 2)         {             if(zeros == -1)             {                 return 1;             }             else             return 0;         }         zeros --;         grid[i][j] = -1;         int total = dfs(grid , i , j+1 , zeros , m , n) +                     dfs(grid , i+1 , j , zeros , m , n) +                       dfs(grid , i , j-1 , zeros , m , n) +                     dfs(grid , i-1 , j , zeros , m , n);         zeros ++;         grid[i][j] = 0;         return total;     }     int uniquePathsIII(vector<vector<int>>& grid) {               int sx =0;               int sy =0 ;               int zeros = 0;               int m  = grid.size();               int n = grid[0].size();               for(int i =0 ; i < m ; i++)               {                   for(int j =0 ; j  < n ; j++)                   {                       if(grid[i][j] == 0)                       {                           zeros++;                       }                       else                       if(grid[i][j] == 1)                       {                           sx = i;                           sy = j;                       }                   }               }               return dfs(grid , sx ,sy , zeros , m , n );     } };