<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:
    
    int helper(string& a,string& b)
    {
        
        unordered_set<string>vis;
        int ans=1,x,n,dif;
        queue<pair<string,int>>q;
        
        n=a.size();
        dif=n;
        
        q.push({a,n});
        vis.insert(a);
        
        while(!q.empty())
        {
            x=q.size();
            
            for(int i=0;i<x;i++)
            {
                for(int j=0;j<n-1;j++)
                {
                    for(int k=j+1;k<n;k++)
                    {
                        if(q.front().first[j]!=q.front().first[k] && q.front().first[j]!=b[j] && q.front().first[k]!=b[k] && (q.front().first[j]==b[k] || q.front().first[k]==b[j]))
                        {
                            swap(q.front().first[j],q.front().first[k]);
                            
                            if(!vis.count(q.front().first))
                            {
                                if(q.front().first==b)
                                    return ans;
                                
                                if(dif>=q.front().second-(q.front().first[j]==b[j])-(q.front().first[k]==b[k]))
                                {
                                    dif=min(dif,q.front().second-(q.front().first[j]==b[j])-(q.front().first[k]==b[k]));
                                    q.push({q.front().first,dif});
                                    vis.insert(q.front().first);
                                }
                            }
                            
                            swap(q.front().first[j],q.front().first[k]);
                        }
                    }
                }
                
                q.pop();
            }
            
            ans++;
        }
        
        return -1;
    }
    
    int kSimilarity(string s1, string s2) {
        
        int n=s1.size();
        string a,b;
        
        for(int i=0;i<n;i++)
        {
            if(s1[i]!=s2[i])
            {
                a.push_back(s1[i]);
                b.push_back(s2[i]);
            }
        }
        
        if(a.empty())
            return 0;
        
        return helper(a,b);
    }
};
62466