980. Unique Paths III

1. 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.

2. Example

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.

3. 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.

4. Solutions

Depth-First Search && Backtracking

m = grid.size(), n = grid.front().size()
Time complexity: O(4$^{mn}$)
Space complexity: O(mn)

class Solution {
public:
    int uniquePathsIII(vector<vector<int>> grid) {
        const int m = grid.size(), n = grid.front().size();
        int to_walk = 0, row = 0, column = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    row = i;
                    column = j;
                } else if (grid[i][j] == 0 || grid[i][j] == 2) {
                    ++to_walk;
                }
            }
        }

        find_path(grid, row, column, to_walk);

        return count;
    }

private:
    int count = 0;

    void find_path(vector<vector<int>> &grid, int row, int column, int to_walk) {
        const int m = grid.size(), n = grid.front().size();
        if (0 <= row && row < m && 0 <= column && column < n) {
            if (grid[row][column] == 2 && to_walk == 0) {
                ++count;
            } else if ((grid[row][column] == 0 || grid[row][column] == 1) && to_walk > 0) {
                grid[row][column] = -1;

                find_path(grid, row + 1, column, to_walk - 1);
                find_path(grid, row - 1, column, to_walk - 1);
                find_path(grid, row, column + 1, to_walk - 1);
                find_path(grid, row, column - 1, to_walk - 1);

                grid[row][column] = 0;
            }
        }
    }
};