<p>Strings <code>s1</code> and <code>s2</code> are <code>k</code><strong>-similar</strong> (for some non-negative integer <code>k</code>) if we can swap the positions of two letters in <code>s1</code> exactly <code>k</code> times so that the resulting string equals <code>s2</code>.</p>

<p>Given two anagrams <code>s1</code> and <code>s2</code>, return the smallest <code>k</code> for which <code>s1</code> and <code>s2</code> are <code>k</code><strong>-similar</strong>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;ba&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The two string are 1-similar because we can use one swap to change s1 to s2: &quot;ab&quot; --&gt; &quot;ba&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;abc&quot;, s2 = &quot;bca&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two strings are 2-similar because we can use two swaps to change s1 to s2: &quot;abc&quot; --&gt; &quot;bac&quot; --&gt; &quot;bca&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length &lt;= 20</code></li>
	<li><code>s2.length == s1.length</code></li>
	<li><code>s1</code> and <code>s2</code> contain only lowercase letters from the set <code>{&#39;a&#39;, &#39;b&#39;, &#39;c&#39;, &#39;d&#39;, &#39;e&#39;, &#39;f&#39;}</code>.</li>
	<li><code>s2</code> is an anagram of <code>s1</code>.</li>
</ul>

class Solution {
public:

    vector<string> find(string &str , string &s2){
        vector<string>ans;
        int i=0;
        while(str[i]==s2[i]) i++;
        for(int j=i;j<str.length();j++){
            if(str[j]==s2[i]){
            swap(str[i] , str[j]);
            ans.push_back(str);
            swap(str[i] , str[j]);
            }
        }
        return ans;
    }


    int kSimilarity(string s1, string s2) {
        queue<string>q;
        unordered_map<string,int>mp;
        mp[s1]=0;
        q.push(s1);
        while(!q.empty()){
            string str=q.front();
            
            if(str==s2) return mp[str];
            q.pop();
            
            for(auto i:find(str , s2)){
                if(mp.find(i)==mp.end()){
                    q.push(i);
                    mp[i]=mp[str]+1;
                }
            }
        }
    return -1;
    }
};
167478