752. Open the Lock LeetCode Solution

In this guide, you will get 752. Open the Lock LeetCode Solution with the best time and space complexity. The solution to Open the Lock problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Open the Lock solution in C++
Open the Lock solution in Java
Open the Lock solution in Python
Additional Resources

752. Open the Lock LeetCode Solution image

Problem Statement of Open the Lock

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.

Complexity Analysis

Time Complexity: O(10^4)
Space Complexity: O(10^4)

752. Open the Lock LeetCode Solution in C++

class Solution {
 public:
  int openLock(vector<string>& deadends, string target) {
    unordered_set<string> seen{deadends.begin(), deadends.end()};
    if (seen.contains("0000"))
      return -1;
    if (target == "0000")
      return 0;

    queue<string> q{{"0000"}};

    for (int step = 1; !q.empty(); ++step)
      for (int sz = q.size(); sz > 0; --sz) {
        string word = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i) {
          const char cache = word[i];
          // Increase the i-th digit by 1.
          word[i] = word[i] == '9' ? '0' : word[i] + 1;
          if (word == target)
            return step;
          if (!seen.contains(word)) {
            q.push(word);
            seen.insert(word);
          }
          word[i] = cache;
          // Decrease the i-th digit by 1.
          word[i] = word[i] == '0' ? '9' : word[i] - 1;
          if (word == target)
            return step;
          if (!seen.contains(word)) {
            q.push(word);
            seen.insert(word);
          }
          word[i] = cache;
        }
      }

    return -1;
  }
};
/* code provided by PROGIEZ */

752. Open the Lock LeetCode Solution in Java

class Solution {
  public int openLock(String[] deadends, String target) {
    Set<String> seen = new HashSet<>(Arrays.asList(deadends));
    if (seen.contains("0000"))
      return -1;
    if (target.equals("0000"))
      return 0;

    Queue<String> q = new ArrayDeque<>(List.of("0000"));

    for (int step = 1; !q.isEmpty(); ++step)
      for (int sz = q.size(); sz > 0; --sz) {
        StringBuilder sb = new StringBuilder(q.poll());
        for (int i = 0; i < 4; ++i) {
          final char cache = sb.charAt(i);
          // Increase the i-th digit by 1.
          sb.setCharAt(i, sb.charAt(i) == '9' ? '0' : (char) (sb.charAt(i) + 1));
          String word = sb.toString();
          if (word.equals(target))
            return step;
          if (!seen.contains(word)) {
            q.offer(word);
            seen.add(word);
          }
          sb.setCharAt(i, cache);
          // Decrease the i-th digit by 1.
          sb.setCharAt(i, sb.charAt(i) == '0' ? '9' : (char) (sb.charAt(i) - 1));
          word = sb.toString();
          if (word.equals(target))
            return step;
          if (!seen.contains(word)) {
            q.offer(word);
            seen.add(word);
          }
          sb.setCharAt(i, cache);
        }
      }

    return -1;
  }
}
// code provided by PROGIEZ