How hard is LeetCode 864. Shortest Path to Get All Keys?

LeetCode 864. Shortest Path to Get All Keys is rated Hard on LeetCode.

What is the time complexity of LeetCode 864. Shortest Path to Get All Keys?

The Python solution on this page runs in O(m \times n \times 2^k).

What is the space complexity of LeetCode 864. Shortest Path to Get All Keys?

The Python solution on this page uses O(m \times n \times 2^k) auxiliary space.

What topics does LeetCode 864. Shortest Path to Get All Keys cover?

LeetCode 864. Shortest Path to Get All Keys is tagged Bit Manipulation, Breadth-First Search, Array and Matrix on LeetCode.

Shortest Path to Get All Keys — LeetCode 864 Python Solution

HardBit ManipulationBreadth-First SearchArrayMatrix

Problem: #864
Pattern: Bit Manipulation
Reading time: 7 min
Source: leetcode.com

The problem

You are given an m x n grid grid where: '.' is an empty cell. '#' is a wall.

Example

Input: grid = ["@.a..","###.#","b.A.B"]
Output: 8
Explanation: Note that the goal is to obtain all the keys not to open all the locks.

Python solution

Python

class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        m, n = len(grid), len(grid[0])
        # Find the starting point (si, sj)
        si, sj = next((i, j) for i in range(m) for j in range(n) if grid[i][j] == '@')
        # Count the number of keys
        k = sum(v.islower() for row in grid for v in row)
        dirs = (-1, 0, 1, 0, -1)
        q = deque([(si, sj, 0)])
        vis = {(si, sj, 0)}
        ans = 0
        while q:
            for _ in range(len(q)):
                i, j, state = q.popleft()
                # If all keys are found, return the current step count
                if state == (1 << k) - 1:
                    return ans

                # Search in the four directions
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    nxt = state
                    # Within boundary limits
                    if 0 <= x < m and 0 <= y < n:
                        c = grid[x][y]
                        # It's a wall, or it's a lock but we don't have the key for it
                        if (
                            c == '#'
                            or c.isupper()
                            and (state & (1 << (ord(c) - ord('A')))) == 0
                        ):
                            continue
                        # It's a key
                        if c.islower():
                            # Update the state
                            nxt |= 1 << (ord(c) - ord('a'))
                        # If this state has not been visited, enqueue it
                        if (x, y, nxt) not in vis:
                            vis.add((x, y, nxt))
                            q.append((x, y, nxt))
            # Increment the step count
            ans += 1
        return -1

Complexity

Measure	Complexity
Time	O(m \times n \times 2^k)
Space	O(m \times n \times 2^k) auxiliary

Pattern: Bit Manipulation

Use XOR, masks and the low-bit trick to replace whole data structures with an integer. LeetCode 864. Shortest Path to Get All Keys is filed here because LeetCode tags it Bit Manipulation, which is the vocabulary this hub collects.

The bit manipulation guide has the Python template for the pattern and the 194 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 864. Shortest Path to Get All Keys?: LeetCode 864. Shortest Path to Get All Keys is rated Hard on LeetCode.
What is the time complexity of LeetCode 864. Shortest Path to Get All Keys?: The Python solution on this page runs in O(m \times n \times 2^k).
What is the space complexity of LeetCode 864. Shortest Path to Get All Keys?: The Python solution on this page uses O(m \times n \times 2^k) auxiliary space.
What topics does LeetCode 864. Shortest Path to Get All Keys cover?: LeetCode 864. Shortest Path to Get All Keys is tagged Bit Manipulation, Breadth-First Search, Array and Matrix on LeetCode.