980. Unique Paths III

Hard (Khó) Python 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Matrix (Ma trận)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

Python 0980-unique-paths-iii.py 
class Solution:
    def uniquePathsIII(self, grid: List[List[int]]) -> int:
        DIR = [1, 0, -1, 0, 1]
        n, m = len(grid), len(grid[0])
        self.cnt = 0
        
        def dfs(x: int, y: int, empty: int) -> None:
            if not (0 <= x < n and 0 <= y < m and grid[x][y] != -1):
                return
            if grid[x][y] == 2:
                self.cnt += empty == 0
                return
            grid[x][y] = -1
            for i in range(4):
                dfs(x + DIR[i], y + DIR[i + 1], empty - 1)
            grid[x][y] = 0
            
        x, y, empty = 0, 0, 1
        
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    x, y = i, j
                elif grid[i][j] == 0:
                    empty += 1
                    
        dfs(x, y, empty)
        return self.cnt