<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 {
private:
    int bfs(string &a, string &b)
    {
        if(a == b) return 0;
		// some setup
        int n = b.size(); // number of chars in string
        int ret = 0; // min swaps to be returned
        unordered_set<string> vis; // visted set to prevent searching strings which have already been searched
        queue<pair<string,pair<int,int>>> q; // the frontier for considering new strings in FIFO order
																	 // stores (string, index, swaps)
		
        q.push(make_pair(b,make_pair(0,0))); // push original string  
        vis.insert(b); // mark as visited
        while(!q.empty())
        {
            auto t = q.front();
            q.pop();
			// get node from front of the queue
            string ss = t.first; // string for current node
            int idx = t.second.first, swaps = t.second.second; // idx: all chars before idx are sorted in ss
																						  // swaps: min number of swaps required to
																						  //             get to reach ss
            
			// increment index till we get to the first char that has to be swapped in current node
            while(a[idx] == ss[idx])
            {
                ++idx;
            }
			// find the appropriate character to swap in ss
            for(int i = idx+1; i < n; ++i)
            {
                if(ss[i] == a[idx])
                {
                    ss[i] = ss[idx];
                    ss[idx] = a[idx];
                    // if the node is equal to target string, return number of swaps required to get to this node
                    if(ss == a) return swaps+1;
					// if not already visited push onto the queue, mark visited
                    if(!vis.count(ss))
                    {
                        vis.insert(ss);
                        q.push(make_pair(ss,make_pair(idx+1,swaps+1)));
                    }
                    ss[idx] = ss[i];
                    ss[i] = a[idx];
                }
            }
        }
        return ret;
    }
public:
    int kSimilarity(string a, string b) 
    {
        return bfs(a, b);;
    }
};
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