Open the Lock

LeetCode Q 752 - 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 can’t reach the target without getting stuck.

Example 4:
Input: deadends = ["0000"], target = "8888" ; Output: -1

Note:

• The length of deadends will be in the range [1, 500].
• target will not be in the list deadends.
• Every string in deadends and the string target will be a string of 4 digits from the 10,000 possibilities ‘0000’ to ‘9999’.

Solution : BFS / Two-End BFS (Bidirectional Search)

Solution 1: BFS

Tip:
Chaning char[] is faster than string concatenation. That is,
char[] chs = cur.toCharArray(); chs[i] = (char)((cur.charAt(i) - '0' + 9) % 10 + '0'); String next = String.valueOf(chs);
is better than
String next = cur.substring(0, i) + (char)((cur.charAt(i) - '0' + 9) % 10 + '0') + cur.substring(i + 1);

Code:

public int openLock(String[] deadends, String target) {
	
	Set<String> deadSet = new HashSet<>();
	for (String str : deadends) 
	    deadSet.add(str);
	
	if (deadSet.contains(target) || deadSet.contains("0000")) return -1;
	
	Queue<String> que = new LinkedList<>();
	Set<String> visited = new HashSet<>();
	que.offer("0000");
	visited.add("0000");
	
	int result = 0;
	while (!que.isEmpty()) {
		
		int size = que.size();
		while (size-- != 0) {
			
			String cur = que.poll();
			
			if (cur.equals(target)) return result;
			
			for (int i = 0; i < 4; ++i) {
				char[] chs = cur.toCharArray();
				
				chs[i] = (char)((cur.charAt(i) - '0' + 9) % 10 + '0');
				String next = String.valueOf(chs);
				if (!visited.contains(next) && !deadSet.contains(next)) {
				    que.offer(next); visited.add(next);
				}
				
				chs[i] = (char)((cur.charAt(i) - '0' + 1) % 10 + '0');
				next = String.valueOf(chs);
				if (!visited.contains(next) && !deadSet.contains(next)) {
				    que.offer(next); visited.add(next);
				}
			}
		}
		
		result++;
	}	
	
	return -1;
}

Solution 2: Two-End BFS (Bidirectional Search)

Code:

public int openLock(String[] deadends, String target) {
	    
	Set<String> deadSet = new HashSet<>();
	for (String str : deadends) deadSet.add(str);
	
	if (deadSet.contains(target) || deadSet.contains("0000")) return -1;
	
	Queue<String> queA = new LinkedList<>();
	Set<String> visitedA = new HashSet<>();
	queA.offer("0000");
	visitedA.add("0000");
	
	Queue<String> queB = new LinkedList<>();
	Set<String> visitedB = new HashSet<>();
	queB.offer(target);
	visitedB.add(target);
	
	Queue<String> que = null; 
	Set<String> setCurr = null; 
	Set<String> setOp = null;
	
	int result = 0;
	while (!queA.isEmpty() && !queB.isEmpty()) {
		if (queA.size() <= queB.size()) {
			que = queA; setCurr = visitedA; setOp = visitedB;
		} else {
			que = queB; setCurr = visitedB; setOp = visitedA;
		}
		
		result++;
		int size = que.size();
		while (size-- != 0) {
			
			String cur = que.poll();
			
			for (int i = 0; i < 4; ++i) {
				char[] chs = cur.toCharArray();
				
				chs[i] = (char)((cur.charAt(i) - '0' + 9) % 10 + '0');
				String next = String.valueOf(chs);
				if (setOp.contains(next)) return result;
				if (!setCurr.contains(next) && !deadSet.contains(next)) {
					que.offer(next); setCurr.add(next);
				}
				
				chs[i] = (char)((cur.charAt(i) - '0' + 1) % 10 + '0');
				next = String.valueOf(chs);
				if (setOp.contains(next)) return result;
				if (!setCurr.contains(next) && !deadSet.contains(next)) {
					que.offer(next); setCurr.add(next);
				}
			}
		}
	}
	
	return -1;
}