Open The Lock Leetcode Problem 752 [Python Solution]

Welcome to another coding session!

Today, we are going to tackle the Open The Lock problem, which is a fascinating puzzle.

This problem falls under the category of graphs and is categorized as a medium difficulty problem on LeetCode.

We will provide a Python solution to this problem.

But before we dive into the code, let's understand the problem and its constraints.

Problem Overview

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, you can turn '9' to '0' or '0' to '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 dead ends, which are combinations that should be avoided.

If the lock reaches any of these dead ends, it cannot be opened.

Your task is to find the minimum total number of turns required to open the lock, or return -1 if it is impossible to open the lock.

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.

Understanding Constraints

Before we jump into solving the problem, it's essential to understand the constraints provided in the problem statement.

1 <= deadends.length <= 500
deadends[i].length == 4
target.length == 4
The target will not be in the list of deadends.
Both the target and deadends consist of digits only.

Now that we have a clear understanding of the problem, its examples, and constraints, let's move on to the solution.

Open The Lock LeetCode Problem Solution

1. Bruteforce Approach

To solve this problem, we can use a Breadth-First Search (BFS) approach to explore all possible combinations while avoiding dead ends.

Here's a Python solution that implements this approach:

View on GitHub ➹

from collections import deque
from typing import List

class Solution:
    def openLock(self, deadends: List[str], target: str) -&gt; int:
        if &quot;0000&quot; in deadends:
            return -1

        def get_children(lock):
            children = []
            for i in range(4):
                digit = str((int(lock[i]) + 1) % 10)
                children.append(lock[:i] + digit + lock[i + 1 :])
                digit = str((int(lock[i]) + 10 - 1) % 10)
                children.append(lock[:i] + digit + lock[i + 1 :])
            return children

        queue = deque()
        visited = set(deadends)
        queue.append((&quot;0000&quot;, 0))  # (lock, turns)

        while queue:
            lock, turns = queue.popleft()
            if lock == target:
                return turns
            for child in get_children(lock):
                if child not in visited:
                    visited.add(child)
                    queue.append((child, turns + 1))
        return -1

2. Efficient Approach with Explanation

This solution efficiently finds the minimum number of turns to open the lock while avoiding dead ends.

Let's break down the code and discuss its logic:

We begin by checking if the initial state, "0000," is in the list of dead ends.

If it is, we return -1 immediately because we cannot proceed from there.

We define a helper function get_children(lock) to generate all possible combinations that can be reached from the current lock state.

This function iterates through each of the four wheels, increments the digit by 1, wraps around to 0 if necessary, and decrements the digit by 1, also considering the wraparound.

We use a queue to perform a Breadth-First Search (BFS) starting from the initial state "0000." Each element in the queue is a tuple consisting of the lock state and the number of turns taken to reach that state.
We maintain a set called visited to keep track of the visited lock combinations.

We add the dead ends to this set to avoid revisiting them.

Inside the BFS loop, we pop the leftmost element from the queue, representing the current lock state and the number of turns taken to reach it.
If the current lock state is equal to the target, we return the number of turns, as we have successfully opened the lock.
For each child state generated by the get_children function, we check if it has not been visited before.

If it hasn't, we add it to the visited set, increment the number of turns, and add it to the queue to explore further.

If we exhaust all possibilities without reaching the target, we return -1, indicating that it's impossible to open the lock.

This efficient approach ensures that we find the shortest path to the target state, thanks to the BFS algorithm.

Now, let's discuss the time and space complexity of this solution.

Time and Space Complexity

The time complexity of this solution is O(10,000), which is the maximum number of possible lock combinations.

In the worst case, we might need to explore all these combinations.

The space complexity is also O(10,000) because we use a queue to store possible combinations, and we maintain a visited set to keep track of visited states.

In both cases, the number of possible combinations limits the space complexity.

Reasoning Behind Our Approach

The key insight in our approach is to use Breadth-First Search (BFS) to explore the possible lock combinations efficiently.

We start from the initial state "0000" and keep generating children states while avoiding dead ends.

This ensures that we find the shortest path to the target state.

Our solution also optimizes the increment and decrement operations for the lock's wheels, ensuring that they wrap around correctly and yield valid states.

Conclusion

In this blog post, we've tackled the Open The Lock problem on LeetCode.

We provided an efficient Python solution that

uses Breadth-First Search (BFS) to find the minimum number of turns required to open the lock while avoiding dead ends.

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For the detailed problem description and to try the code yourself, you can visit the LeetCode problem page: Open The Lock LeetCode Problem.

Happy coding and keep exploring new problems and their solutions!