What is the time complexity of Open the Lock?

The optimal solution has a time complexity of O(10000).

What is the space complexity of Open the Lock?

The space complexity is O(10000).

What data structures are used to solve Open the Lock?

This problem uses Array, Hash Table, String, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Open the Lock?

Insight 1: Breadth-First Search (BFS) is the optimal algorithm for finding the shortest path in an unweighted graph. Each lock state can be thought of as a node, and each possible move (turning a wheel) is an edge. Insight 2: The lock states can be represented as strings and stored in a set or hash table to quickly check for deadends and visited states.

Which companies ask LeetCode 752. Open the Lock in interviews?

This problem is frequently asked at ByteDance, CARS24, Citadel, Coupang, DE Shaw and 4 more companies.

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Open the Lock - LeetCode 752 Solution

Open the Lock - Complete Solution Guide

Open the Lock is LeetCode problem 752, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the minimum number of moves (turns of the wheels) required to transform a 4-digit lock from its initial state '0000' to a given 'target' state. We are given a list of 'deadends', which represent combinations that, if encountered, prevent the lock from being opened. The lock has 4 wheels, each with digits 0-9, and each move consists of turning one wheel one digit forward or backward (wrapping around, so 9 goes to 0 and 0 goes to 9). The target will not be in the deadends, and the target and deadends will only contain digits.

Solution Approach

The provided code utilizes Bidirectional Breadth-First Search (BFS) to find the shortest path from the initial state '0000' to the target state while avoiding deadends. Two queues are used, one starting from '0000' and the other starting from 'target'. At each step, the algorithm expands the queue with fewer elements to minimize the search space. The algorithm explores all possible neighbor states by incrementing or decrementing each digit. If a neighbor state is found in the visited states of the other queue, it means a path has been found, and the sum of the distances from both sides gives the result. If the target is not reachable without hitting a deadend, the function returns -1.

Step-by-Step Algorithm

Step 1: Initialize a set of deadends for fast lookup.
Step 2: Check if the initial state '0000' is a deadend. If so, return -1 because the lock cannot be opened.
Step 3: Check if the target is '0000'. If so, return 0 because no moves are needed.
Step 4: Initialize two queues, q1 starting at '0000' and q2 starting at the 'target'.
Step 5: Initialize two hash maps (or sets), visited1 and visited2, to keep track of the visited states and their distances from their respective starting points.
Step 6: While both queues are not empty, expand the smaller queue.
Step 7: For each state in the current queue, generate all possible neighbor states by incrementing/decrementing each digit.
Step 8: For each neighbor state, check if it has been visited by the other search. If so, a path has been found; return the combined distance.
Step 9: If the neighbor state is not a deadend and has not been visited by the current search, add it to the queue and mark it as visited.
Step 10: If both queues become empty without finding a path, it implies that the target is unreachable; return -1.

Key Insights

Insight 1: Breadth-First Search (BFS) is the optimal algorithm for finding the shortest path in an unweighted graph. Each lock state can be thought of as a node, and each possible move (turning a wheel) is an edge.
Insight 2: The lock states can be represented as strings and stored in a set or hash table to quickly check for deadends and visited states.
Insight 3: Bidirectional BFS can significantly improve performance by searching from both the start and the end simultaneously. This reduces the search space.

Complexity Analysis

Time Complexity: O(10000)

Space Complexity: O(10000)

Topics

This problem involves: Array, Hash Table, String, Breadth-First Search.

Companies

Asked at: ByteDance, CARS24, Citadel, Coupang, DE Shaw, Flipkart, J.P. Morgan, Zepto, eBay.

Open the Lock | LeetCode 752

Medium

Array Hash Table String Breadth-First Search

Time: O(10000)Space: O(10000)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int openLock(String[] deadends, String target) {
        Set<String> dead_set = new HashSet<>(Arrays.asList(deadends));
        
        if (dead_set.contains("0000")) {
            return -1;
        }
        
        if (target.equals("0000")) {
            return 0;
        }
        
        // Bidirectional BFS
        Queue<String> q1 = new LinkedList<>(Collections.singletonList("0000"));
        Queue<String> q2 = new LinkedList<>(Collections.singletonList(target));
        
        Map<String, Integer> visited1 = new HashMap<>();
        visited1.put("0000", 0);
        Map<String, Integer> visited2 = new HashMap<>();
        visited2.put(target, 0);
        
        while (!q1.isEmpty() && !q2.isEmpty()) {
            // Heuristic to expand the smaller queue first
            if (q1.size() > q2.size()) {
                Queue<String> tempQ = q1;
                q1 = q2;
                q2 = tempQ;

                Map<String, Integer> tempMap = visited1;
                visited1 = visited2;
                visited2 = tempMap;
            }

            int size = q1.size();
            for (int _i = 0; _i < size; _i++) {
                String current_state = q1.poll();
                int moves = visited1.get(current_state);
                
                for (int i = 0; i < 4; i++) {
                    int digit = current_state.charAt(i) - '0';
                    for (int d : new int[]{-1, 1}) {
                        // Rotate wheel i by d slots
                        int next_digit = (digit + d + 10) % 10;
                        String neighbor_state = current_state.substring(0, i) + String.valueOf(next_digit) + current_state.substring(i+1);

                        // Check if the two searches meet
                        if (visited2.containsKey(neighbor_state)) {
                            return moves + 1 + visited2.get(neighbor_state);
                        }

                        // Add new state to the queue if not visited and not a deadend
                        if (!visited1.containsKey(neighbor_state) && !dead_set.contains(neighbor_state)) {
                            visited1.put(neighbor_state, moves + 1);
                            q1.offer(neighbor_state);
                        }
                    }
                }
            }
        }
        
        return -1;
    }
}

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Problem Info

Description

Problem 752: Open the Lock - Detailed explanation and solution approaches.

Topics

Array Hash Table String Breadth-First Search

Companies

ByteDanceCARS24CitadelCoupangDE ShawFlipkartJ.P. MorganZepto

Stats

Difficulty:Medium

Languages:4

Companies:9

Problem #:752

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a set of deadends for fast lookup.
2
Step 2: Check if the initial state '0000' is a deadend. If so, return -1 because the lock cannot be opened.
3
Step 3: Check if the target is '0000'. If so, return 0 because no moves are needed.
4
Step 4: Initialize two queues, q1 starting at '0000' and q2 starting at the 'target'.
5
Step 5: Initialize two hash maps (or sets), visited1 and visited2, to keep track of the visited states and their distances from their respective starting points.
6
Step 6: While both queues are not empty, expand the smaller queue.
7
Step 7: For each state in the current queue, generate all possible neighbor states by incrementing/decrementing each digit.
8
Step 8: For each neighbor state, check if it has been visited by the other search. If so, a path has been found; return the combined distance.
9
Step 9: If the neighbor state is not a deadend and has not been visited by the current search, add it to the queue and mark it as visited.
10
Step 10: If both queues become empty without finding a path, it implies that the target is unreachable; return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(10000)

💾

Space Complexity

O(10000)

📝

Why?

There are 10,000 possible states for the lock (0000 to 9999). In the worst-case scenario, we might have to visit each of these states. Both the time and space complexity are proportional to the number of possible states. The bidirectional BFS can, in practice, significantly reduce the search space but doesn't change the theoretical worst-case complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Unidirectional BFS

A standard BFS starting from '0000' and searching for the target state.

Time: O(10000)Space: O(10000)

Trade-offs: Simpler to implement, but potentially slower than bidirectional BFS, especially when the target is far from the start.

A* Search

Using a heuristic function (e.g., the number of digits that differ between the current state and the target) to guide the search. However, in this problem, A* is generally not better than BFS because the search space is relatively uniform and the heuristic is not very informative.

Time: O(b^d) in the worst case, where b is the branching factor and d is the depth.Space: O(b^d)

Trade-offs: Potentially better if a highly accurate heuristic can be designed, but BFS is usually preferred for its simplicity and comparable performance in this specific scenario.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is the optimal algorithm for finding the shortest path in an unweighted graph. Each lock state can be thought of as a node, and each possible move (turning a wheel) is an edge.
2Insight 2: The lock states can be represented as strings and stored in a set or hash table to quickly check for deadends and visited states.
3Insight 3: Bidirectional BFS can significantly improve performance by searching from both the start and the end simultaneously. This reduces the search space.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: The initial state '0000' is a deadend. The solution handles this by checking at the start and returning -1.
Edge case 2: The target state is the same as the initial state '0000'. The solution handles this by returning 0 at the start.
Edge case 3: The target state is unreachable because all paths lead to a deadend. The solution handles this by returning -1 after both BFS queues are empty.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the initial state is a deadend. This will lead to an infinite loop or incorrect result.

✗Common mistake 2: Not using a set/hash table to keep track of visited states. This can lead to revisiting states and incorrect distance calculations.

✗Common mistake 3: Not correctly handling the wrapping around of digits (0 to 9 and 9 to 0).

✗Common mistake 4: Forgetting to add the 'moves' to the neighboring states when adding them to the queue. Each added neighbor state's moves should be the 'current_state's' moves + 1.

✗Common mistake 5: Not understanding the benefit of Bidirectional Search and just implementing regular BFS, which is less efficient in many cases.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Password cracking: Trying different combinations of characters to unlock an account. The 'deadends' can represent known common passwords or patterns to avoid.

🏦Real-world scenario 2: Pathfinding in a maze or game environment: The lock states can be analogous to locations in a maze, and the 'deadends' can be obstacles or blocked paths. The goal is to find the shortest path to a target location.

🎮Real-world scenario 3: State space search in AI: Many AI problems can be framed as searching through a state space, similar to the lock problem. For example, planning a sequence of actions for a robot or solving a puzzle like the Rubik's Cube.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Open the Lock"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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