980

980. Unique Paths III (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.

Companies:
Microsoft, Cruise Automation, Apple

Related Topics:
Array, Backtracking, Bit Manipulation, Matrix

Similar Questions:

• Sudoku Solver (Hard)
• Unique Paths II (Medium)
• Word Search II (Hard)

Solution 1. DFS

// OJ: https://leetcode.com/problems/unique-paths-iii/
// Author: github.com/lzl124631x
// Time: O(3^(MN))
// Space: O(MN) due to call stack
class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), sx, sy, goal = 0, ans = 0, dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
        function<void(int, int)> dfs = [&](int x, int y) {
            if (A[x][y] == 2) {
                ans += goal == 1;
                return;
            }
            A[x][y] = -1;
            --goal;
            for (auto &[dx, dy] : dirs) {
                int a = x + dx, b = y + dy;
                if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] == -1) continue;
                dfs(a, b);
            }
            ++goal;
            A[x][y] = 0;
        };
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] != -1) ++goal;
                if (A[i][j] == 1) sx = i, sy = j;
            }
        }
        dfs(sx, sy);
        return ans;
    }
};

Or we are not allowed to change the grid, we can use a bitmask to track visited cells.

// OJ: https://leetcode.com/problems/unique-paths-iii/
// Author: github.com/lzl124631x
// Time: O(3^(MN))
// Space: O(MN)
class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), sx, sy, goal = 0, ans = 0, dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}}, seen = 0;
        function<void(int, int)> dfs = [&](int x, int y) {
            if (A[x][y] == 2) {
                ans += goal == 1;
                return;
            }
            int bit = 1 << (x * N + y);
            seen ^= bit;
            --goal;
            for (auto &[dx, dy] : dirs) {
                int a = x + dx, b = y + dy;
                if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] == -1 || seen & (1 << (a * N + b))) continue;
                dfs(a, b);
            }
            ++goal;
            seen ^= bit;
        };
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] != -1) ++goal;
                if (A[i][j] == 1) sx = i, sy = j;
            }
        }
        dfs(sx, sy);
        return ans;
    }
};