<p>You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: <code>&#39;0&#39;, &#39;1&#39;, &#39;2&#39;, &#39;3&#39;, &#39;4&#39;, &#39;5&#39;, &#39;6&#39;, &#39;7&#39;, &#39;8&#39;, &#39;9&#39;</code>. The wheels can rotate freely and wrap around: for example we can turn <code>&#39;9&#39;</code> to be <code>&#39;0&#39;</code>, or <code>&#39;0&#39;</code> to be <code>&#39;9&#39;</code>. Each move consists of turning one wheel one slot.</p>

<p>The lock initially starts at <code>&#39;0000&#39;</code>, a string representing the state of the 4 wheels.</p>

<p>You are given a list of <code>deadends</code> 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.</p>

<p>Given a <code>target</code> 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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> deadends = [&quot;0201&quot;,&quot;0101&quot;,&quot;0102&quot;,&quot;1212&quot;,&quot;2002&quot;], target = &quot;0202&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
A sequence of valid moves would be &quot;0000&quot; -&gt; &quot;1000&quot; -&gt; &quot;1100&quot; -&gt; &quot;1200&quot; -&gt; &quot;1201&quot; -&gt; &quot;1202&quot; -&gt; &quot;0202&quot;.
Note that a sequence like &quot;0000&quot; -&gt; &quot;0001&quot; -&gt; &quot;0002&quot; -&gt; &quot;0102&quot; -&gt; &quot;0202&quot; would be invalid,
because the wheels of the lock become stuck after the display becomes the dead end &quot;0102&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> deadends = [&quot;8888&quot;], target = &quot;0009&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We can turn the last wheel in reverse to move from &quot;0000&quot; -&gt; &quot;0009&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> deadends = [&quot;8887&quot;,&quot;8889&quot;,&quot;8878&quot;,&quot;8898&quot;,&quot;8788&quot;,&quot;8988&quot;,&quot;7888&quot;,&quot;9888&quot;], target = &quot;8888&quot;
<strong>Output:</strong> -1
<strong>Explanation:</strong> We cannot reach the target without getting stuck.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= deadends.length &lt;= 500</code></li>
	<li><code>deadends[i].length == 4</code></li>
	<li><code>target.length == 4</code></li>
	<li>target <strong>will not be</strong> in the list <code>deadends</code>.</li>
	<li><code>target</code> and <code>deadends[i]</code> consist of digits only.</li>
</ul>

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        set<string>st;
        set<string>visited;
        for(auto x:deadends)st.insert(x);
        // int steps=0;
        string code="0000";
        if(st.find(code)!=st.end()) return -1;
        queue<pair<string,int>>q; //the string and the strp which we are to find 
        q.push({code,0});
        // visited.insert(code);
        while(q.size()){
            string s=q.front().first;
            int stp=q.front().second;
            q.pop();
            if(visited.find(s)!=visited.end()||st.find(s)!=st.end()) continue;//this one is banned or visited
            visited.insert(s);
            if(s==target){
               return  stp; //this one should hit 
            }else{
                for(int i=0;i<4;i++){
                    int no=s[i]-48;
                    string dc="";
                    string inc="";
                    int dcval,incval;
                    if(no==0){
                        dcval=9;
                        incval=1;
                    }else if(no==9){
                        dcval=8;
                        incval=0;
                    }else{
                        dcval=no-1;
                        incval=no+1;
                    }
                    for(int k=0;k<4;k++){
                        if(k==i){
                            dc+=to_string(dcval);
                            inc+=to_string(incval);
                        }else{
                            dc+=s[k];
                            inc+=s[k];
                        }
                    }
                    //by no we have our new two sting avaiable for i idx of s
                    q.push({dc,stp+1});
                    q.push({inc,stp+1});
                }
            }

        }
        return -1;
    }
};
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