<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 kSimilarity(string s1, string s2) {
        
        unordered_map<string,int> vis;
        queue<string> q;

        q.push(s1);
        vis[s1]=1;
        string curr;
        int steps=0;
        while(!q.empty())
        {
            int z=q.size();

            while(z--)
            {
                curr=q.front();
                q.pop();

                if(curr==s2)
                    return steps;
                
                int i=0;
                while(i<curr.size() && curr[i]==s2[i])
                    i++;

                for(int l=i+1;l<curr.size();l++)
                {
                    if(curr[l]==s2[i])
                    {
                        string nn=curr;
                        nn[l]=curr[i];
                        nn[i]=curr[l];

                        if(vis.find(nn)==vis.end())
                        {
                            q.push(nn);
                            vis[nn]=1;
                        }
                    }
                }
            }

            steps++;
        }

        return steps;
    }

};

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