Introduction

In this blog post, we will explore how to solve the LeetCode problem "864" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n grid grid where: '.' is an empty cell. '#' is a wall. '@' is the starting point. Lowercase letters represent keys. Uppercase letters represent locks. You start at the starting point and one move consists of walking one space in one of the four cardinal directions. You cannot walk outside the grid, or walk into a wall. If you walk over a key, you can pick it up and you cannot walk over a lock unless you have its corresponding key. For some 1 <= k <= 6, there is exactly one lowercase and one uppercase letter of the first k letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet. Return the lowest number of moves to acquire all keys. If it is impossible, return -1. Example 1: 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. Example 2: Input: grid = ["@..aA","..B#.","....b"] Output: 6 Example 3: Input: grid = ["@Aa"] Output: -1 Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 30 grid[i][j] is either an English letter, '.', '#', or '@'. There is exactly one '@' in the grid. The number of keys in the grid is in the range [1, 6]. Each key in the grid is unique. Each key in the grid has a matching lock.

Explanation

Here's the breakdown of the solution:

• BFS with State Representation: Use Breadth-First Search (BFS) to explore the grid. The state at each step includes the current row, column, and the keys collected so far (represented as a bitmask).
• Key Tracking (Bitmask): Employ a bitmask to efficiently represent which keys have been collected. Each key corresponds to a specific bit in the bitmask (e.g., 'a' is bit 0, 'b' is bit 1, etc.).

• Visited Set Optimization: Utilize a set to keep track of visited states (row, column, keys) to avoid revisiting the same state, which optimizes performance by pruning redundant paths.

• Runtime Complexity: O(m n 2^k) where m is rows, n is cols, and k is the number of keys.

• Space Complexity: O(m n 2^k)

Code

    from collections import deque

def shortestPathAllKeys(grid):
    m, n = len(grid), len(grid[0])
    start_row, start_col = -1, -1
    num_keys = 0
    for i in range(m):
        for j in range(n):
            if grid[i][j] == '@':
                start_row, start_col = i, j
            elif 'a' <= grid[i][j] <= 'f':
                num_keys += 1

    queue = deque([(start_row, start_col, 0, 0)])  # row, col, keys_bitmask, moves
    visited = set()
    visited.add((start_row, start_col, 0))

    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while queue:
        row, col, keys, moves = queue.popleft()

        if keys == (1 << num_keys) - 1:
            return moves

        for dr, dc in directions:
            new_row, new_col = row + dr, col + dc

            if 0 <= new_row < m and 0 <= new_col < n and grid[new_row][new_col] != '#':
                cell = grid[new_row][new_col]

                if 'a' <= cell <= 'f':
                    new_keys = keys | (1 << (ord(cell) - ord('a')))
                    if (new_row, new_col, new_keys) not in visited:
                        visited.add((new_row, new_col, new_keys))
                        queue.append((new_row, new_col, new_keys, moves + 1))
                elif 'A' <= cell <= 'F':
                    lock_bit = 1 << (ord(cell) - ord('A'))
                    if (keys & lock_bit) != 0:
                        if (new_row, new_col, keys) not in visited:
                            visited.add((new_row, new_col, keys))
                            queue.append((new_row, new_col, keys, moves + 1))
                else:
                    if (new_row, new_col, keys) not in visited:
                        visited.add((new_row, new_col, keys))
                        queue.append((new_row, new_col, keys, moves + 1))

    return -1