<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;
        
        queue<pair<string,int>> q;

        unordered_set<string> seen;
        int i =0;
        while(i<s1.size() && s1[i]==s2[i]){
            i++;
        }

        for(int j =i+1 ; j<s1.size() ; j++){

            if(s2[i] == s1[j]){
                string temp = s1;
                swap(temp[i],temp[j]);
                seen.insert(temp);
                q.push({temp,1});
            }

        }


        while(!q.empty()){

            auto fNode = q.front();
            q.pop();
            int level = fNode.second;
            if(fNode.first == s2){
                return fNode.second;
            }
            int i =0 ;
            while(i<s1.size() && fNode.first[i]==s2[i]){
            i++;
            }

            for(int j =i+1 ; j<s1.size() ; j++){

                    if(s2[i] == fNode.first[j]){
                        string temp = fNode.first;
                        swap(temp[i],temp[j]);
                        if(seen.find(temp)==seen.end()){
                             seen.insert(temp);
                             q.push({temp,level+1});
                        }
                    }

            }


        }


        return 0;

    }
};
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