Introduction

In this blog post, we will explore how to solve the LeetCode problem "752" 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 have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot. The lock initially starts at '0000', a string representing the state of the 4 wheels. You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it. Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible. Example 1: Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202" Output: 6 Explanation: A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202". Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid, because the wheels of the lock become stuck after the display becomes the dead end "0102". Example 2: Input: deadends = ["8888"], target = "0009" Output: 1 Explanation: We can turn the last wheel in reverse to move from "0000" -> "0009". Example 3: Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888" Output: -1 Explanation: We cannot reach the target without getting stuck. Constraints: 1 <= deadends.length <= 500 deadends[i].length == 4 target.length == 4 target will not be in the list deadends. target and deadends[i] consist of digits only.

Explanation

• Breadth-First Search (BFS): Use BFS to explore the lock's state space, starting from "0000". Each state represents a possible lock configuration.
  • Deadend Avoidance: Keep track of visited states and deadends to avoid cycles and invalid states.
  • State Generation: For each state, generate all possible next states by turning each wheel one slot forward or backward.

• Runtime Complexity: O(10000), where 10000 is the size of the state space (all possible 4-digit combinations). Storage Complexity: O(10000) in the worst case, to store the visited set and queue.

Code

    from collections import deque

def open_lock(deadends, target):
    """
    Finds the minimum number of turns required to open the lock.

    Args:
        deadends: A list of deadend strings.
        target: The target string.

    Returns:
        The minimum number of turns, or -1 if impossible.
    """

    deadends_set = set(deadends)
    if "0000" in deadends_set:
        return -1

    queue = deque([("0000", 0)])  # (state, moves)
    visited = {"0000"}

    while queue:
        state, moves = queue.popleft()

        if state == target:
            return moves

        for i in range(4):
            digit = int(state[i])

            # Turn forward
            new_digit = (digit + 1) % 10
            new_state = state[:i] + str(new_digit) + state[i+1:]
            if new_state not in deadends_set and new_state not in visited:
                queue.append((new_state, moves + 1))
                visited.add(new_state)

            # Turn backward
            new_digit = (digit - 1) % 10
            new_state = state[:i] + str(new_digit) + state[i+1:]
            if new_state not in deadends_set and new_state not in visited:
                queue.append((new_state, moves + 1))
                visited.add(new_state)

    return -1