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# Open the Lock LeetCode Solution – Queslers

## Problem – 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.

### Open the Lock LeetCode Solution in Python

``````    def openLock(self, deadends, target):
marker, depth = 'x', -1

while q:
size = len(q)
depth += 1
for _ in range(size):
node = q.popleft()
if node == target: return depth
if node in visited: continue
q.extend(self.successors(node))
return -1

def successors(self, src):
res = []
for i, ch in enumerate(src):
num = int(ch)
res.append(src[:i] + str((num - 1) % 10) + src[i+1:])
res.append(src[:i] + str((num + 1) % 10) + src[i+1:])
return res
``````

### Open the Lock LeetCode Solution in Java

``````public static int openLock(String[] deadends, String target) {
int depth = -1;
while(!q.isEmpty()) {
depth++;
int size = q.size();
for(int i = 0; i < size; i++) {
String node = q.poll();
if(node.equals(target))
return depth;
if(visited.contains(node))
continue;
}
}
return -1;
}

private static List<String> getSuccessors(String str) {
for (int i = 0; i < str.length(); i++) {
res.add(str.substring(0, i) + (str.charAt(i) == '0' ? 9 :  str.charAt(i) - '0' - 1) + str.substring(i+1));
res.add(str.substring(0, i) + (str.charAt(i) == '9' ? 0 :  str.charAt(i) - '0' + 1) + str.substring(i+1));
}
return res;
}
``````

### Open the Lock LeetCode Solution in C++

``````class Solution {
public:
int openLock(vector<string>& deadends, string target) {
unordered_set<string> visited;
queue<string> bfs;
string init = "0000";
if (dds.find(init) != dds.end()) return -1;
visited.insert("0000");
bfs.push("0000");
int res = 0;
while (!bfs.empty()) {
int sz = bfs.size();
for (int i = 0; i < sz; i++) {
string t = bfs.front(); bfs.pop();
vector<string> nbrs = move(nbrStrs(t));
for (auto s : nbrs) {
if (s == target) return ++res;
if (visited.find(s) != visited.end()) continue;
if (dds.find(s) == dds.end()) {
bfs.push(s);
visited.insert(s);
}
}
}
++res;
}
return -1;
}

vector<string> nbrStrs(string key) {
vector<string> res;
for (int i = 0 ; i < 4; i++) {
string tmp = key;
tmp[i] = (key[i] - '0' + 1) % 10 + '0';
res.push_back(tmp);
tmp[i] = (key[i] - '0' - 1 + 10) % 10 + '0';
res.push_back(tmp);
}
return res;
}
};
``````
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