Open the Lock Problem

Problem

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.

Examples

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.

Solution

Method 1 - BFS

Code

public int openLock(String[] deadends, String target) {
	Queue<String> q = new LinkedList<>();
	Set<String> deads = new HashSet<>(Arrays.asList(deadends));
	Set<String> visited = new HashSet<>();
	q.offer("0000");
	visited.add("0000");
	int level = 0;
	while(!q.isEmpty()) {
		int size = q.size();
		while(size > 0) {
			String s = q.poll();
			if(deads.contains(s)) {
				size --;
				continue;
			}
			if(s.equals(target)) {
				return level;
			}
			StringBuilder sb = new StringBuilder(s);
			for(int i = 0; i < 4; i ++) {
				char c = sb.charAt(i);
				String s1 = sb.substring(0, i) + (c == '9' ? 0 : c - '0' + 1) + sb.substring(i + 1);
				String s2 = sb.substring(0, i) + (c == '0' ? 9 : c - '0' - 1) + sb.substring(i + 1);
				if(!visited.contains(s1) && !deads.contains(s1)) {
					q.offer(s1);
					visited.add(s1);
				}
				if(!visited.contains(s2) && !deads.contains(s2)) {
					q.offer(s2);
					visited.add(s2);
				}
			}
			size --;
		}
		level ++;
	}
	return -1;
}

We can also replace check for char 9 and 0 by mod operator:

String s1 = sb.substring(0, i) + (c - '0' + 1)%10 + sb.substring(i + 1);

WE can also simplify code further by using deadends as part of visited set. Also, we can get all the state from current string as a separate function:

public int openLock(String[] deadends, String target) {
	if (target == null || target.length() == 0) return -1;
	Set<String> visited = new HashSet<>(Arrays.asList(deadends));
	Queue<String> queue = new LinkedList<>();
	int level = 0;
	queue.offer("0000");
	
	while (!queue.isEmpty()) {
		int size = queue.size();
		for (int i = 0; i < size; i++) {
			String currentLock = queue.poll();
			if (!visited.add(currentLock)) {
				continue;
			}
			if (currentLock.equals(target)) {
					return level;
				}
			
			for (String nextLock : getNextStates(currentLock)) {
				if (!visited.contains(nextLock)) {
					queue.offer(nextLock);
				}
			}
		}
		level++;
	}
	
	return -1;
}

private List<String> getNextStates(String lock) {
	List<String> locks = new LinkedList<>();
	char[] arr = lock.toCharArray();
	for (int i = 0; i < arr.length; i++) {
		char c = arr[i];
		arr[i] = (c - '0' + 1) % 10;
		locks.add(String.valueOf(arr));
		arr[i] = c == (c - '0' - 1) % 10;
		locks.add(String.valueOf(arr));
		arr[i] = c;
	}
	return locks;
}

Open the Lock Problem#

Problem#

Examples#

Solution#

Method 1 - BFS#

Code#