Unique Paths III | Leetcode 980 Solution

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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 );
    }
};

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