<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) {
        queue<pair<string,vector<int>>> q;
        q.push({s1,{0,0}});
        unordered_set<string> visited;
        while(!q.empty()){
            int curr_pos = q.front().second[0];
            int curr_depth = q.front().second[1];
            string s = q.front().first;
            q.pop();
            if(s == s2) return curr_depth;
            if (curr_pos >= s.length()) continue;
            while(curr_pos < s.length() && s[curr_pos] == s2[curr_pos]){
                curr_pos++;
            }
            for(int i = curr_pos + 1; i < s.length(); i++){
                if(s[i] == s2[curr_pos]){
                    
                    swap(s[curr_pos], s[i]);
                    if(!visited.contains(s)){
                        visited.insert(s);
                        q.push({s,{curr_pos+1,curr_depth+1}});
                    }
                    swap(s[i], s[curr_pos]);
                }
            }


        }
        return -1;
    }
};
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