<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:
    vector<string> next(string &s,string &s2){
        int i=0;
        while( i<s.size() && s[i]==s2[i])i++;
        vector<string>n;
        for(int j=i+1;j<s.size();++j){
            if(s[j]==s2[i] && s[j]!=s2[j]){
                swap(s[j],s[i]);
                n.push_back(s);
                swap(s[i],s[j]);
            }
        }
        return n;
    }
    int kSimilarity(string s1, string s2) {
        unordered_set<string>st{{s1}};
        queue<string>q;
        q.push(s1);
        int ans = 0;
        while(q.size()){
            int sz = q.size();
            for(int i=0;i<sz;++i){
                string s = q.front();
                q.pop();
                if(s==s2)return ans;
                vector<string>str = next(s,s2);
                for(auto x:str){
                    if(st.find(x)==st.end()){
                        st.insert(x);
                        q.push(x);
                    }
                }
            }
            ans++;
        }
        return ans;
    }
};
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