Unique Paths III

Description

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.

Code

class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int startI, startJ;
        int n = grid.size();
        int m = grid[0].size();
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == 1) {
                    startI = i;
                    startJ = j;
                    break; 
                }
            }
        }
        return search(startI, startJ, grid);
    }
 
    int search(int x, int y, vector<vector<int>>& grid) {
        if(x >= grid.size() || x < 0 || y >= grid[0].size() || y < 0) return 0; 
        if(grid[x][y] == 3 || grid[x][y] == -1) return 0;
        if(grid[x][y] == 2) {
            for(int i = 0; i < grid.size(); i++) {
                for(int j = 0; j < grid[i].size(); j++) {
                    if(grid[i][j] == 0) return 0; 
                }
            }
            return 1;
        }
	    // Backtracking
        int temp = grid[x][y];
        grid[x][y] = 3;
        int res = search(x + 1, y, grid) + search(x - 1, y, grid) + search(x, y + 1, grid) + search(x, y - 1, grid);       
        // Backtracking
        grid[x][y] = temp;
        return res;
 
 
    }
};

Source

• Unique Paths III - LeetCode