<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 A, string B) {
        if(A == B) return 0;
        
        unordered_set<string> visit;
        queue<pair<string, int>> q;
        
        int n = A.size();
        int step = 0;
        
        int i = 0;
        for(; i < n; ++i) {
            if(A[i] != B[i]) break;
        }
        
        q.push({A, i});
        visit.insert(A);
        
        while(!q.empty()) {
            int size = q.size();
            
            for(int i = 0; i < size; ++i) {
                string t = move(q.front().first);
                int index = q.front().second;
                q.pop();
                
                if(t == B) return step;
                
                while(t[index] == B[index]) ++index;
                for(int i = index +1; i < n; ++i) {
                    if(t[i] == B[index] && t[i] != B[i]) {
                        swap(t[index], t[i]);
                        
                        if(visit.count(t) == 0) {
                            q.push({t, index+1});
                            visit.insert(t);
                        }
                        
                        swap(t[index], t[i]);
                    }
                }
                
            }
            
            ++step;
        }
        
        return step;
    }
};
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