<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.size() != s2.size())
            return false;
        int n = s1.size();
        queue<string>to_visit;
        to_visit.push(s2);
        unordered_set<string>visited;
        visited.insert(s2);
        int lvl = 0;
        while(!to_visit.empty()){
            auto s = to_visit.size();
            while(s--){
                auto cur = to_visit.front();
                to_visit.pop();
                int i = 0;
                
                for(;i<n;i++){
                    if(s1[i] != cur[i])
                        break;
                }
                if(i == n)
                    return lvl;
                for(int j = i+1;j<n;j++){
                    if(cur[j] == s1[i]){
                        string nex = cur;
                        swap(nex[i], nex[j]);
                        if(visited.insert(nex).second)
                            to_visit.push(nex);
                    }
                }
            }
            lvl++;
        }
        return -1;
    }
};
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