<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) {
        if(s1 == s2)
        {
            return 0;
        }

        int minMoves = 0;

        int length = s1.length();
        string S1 = "", S2 = "";

        for(int i = 0; i < s1.length(); i++)
        {
            if(s1[i] != s2[i])
            {
                for(int j = i + 1; j < s2.length(); j++)   
                {
                    if(s1[i] == s2[j] && s1[j] == s2[i])
                    {
                        swap(s2[i], s2[j]);
                        minMoves++;
                        break;
                    }
                }
            }

            if(s1[i] != s2[i])    
            {
                S1 += s1[i];
                S2 += s2[i];
            }
        }

        if(S1 == S2)   
        {
            return minMoves;
        }


        length = S1.length();
        queue<pair<string, int>> BFS;
        BFS.push({S2, 0});          

        while(!BFS.empty())
        {
            int size = BFS.size();
            minMoves++;

            for(int i = 0; i < size; i++)
            {
                string val = BFS.front().first;
                int id = BFS.front().second;
                BFS.pop();

                for(int j = id + 1; j < length; j++)
                {
                    if(val[j] == S1[id])
                    {
                        swap(val[id], val[j]);
                        if(val == S1)
                        {
                            return minMoves;
                        }
                        int newID = id + 1;
                        while(val[newID] == S1[newID])
                        {
                            newID++;
                        }
                        BFS.push({val, newID});
                        swap(val[id], val[j]);
                    }
                }
            }
        }
        return minMoves;
    }
};
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