<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 n=s1.size();
        unordered_set<string> mp;
        queue<pair<string,int>> q;
        int index=0;
        for(int i=0;i<n;i++){
            if(s1[i]!=s1[i]){
                index=i;
                break;
            }
        }
        q.push({s1,index});
        int ans=0;
        while(!q.empty()){
            int len=q.size();
            for(int k=0;k<len;k++){
                string s=q.front().first;
                int curr=q.front().second;
                q.pop();
                if(s==s2){
                    return ans;
                }
                while(s[curr]==s2[curr]){
                    curr++;
                }
                for(int i=curr+1;i<n;i++){
                    if(s[i]==s2[i]){
                        continue;
                    }
                    if(s2[curr]==s[i]&&s[i]!=s2[i]){
                        swap(s[i],s[curr]);
                        if(mp.count(s)==0){
                            mp.insert(s);
                            q.push({s,1+curr});
                        }
                        swap(s[i],s[curr]);
                    }
                }
            }
            ans++;
        }
        return 0;
    }
};
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