<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>

from collections import deque

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        # Check if s1 is already the same as s2
        if s1 == s2:
            return 0
        
        # Initialize BFS
        queue = deque([s1])  # Queue holds strings to be processed
        visited = set([s1])  # Set to keep track of visited strings
        swaps = 0  # Number of swaps performed
        
        while queue:
            size = len(queue)
            for _ in range(size):
                current = queue.popleft()
                
                # Generate all possible swaps
                if current == s2:
                    return swaps
                
                # Find the first position where s1 and s2 differ
                i = 0
                while i < len(current) and current[i] == s2[i]:
                    i += 1
                
                # Perform swaps only from position i onwards
                for j in range(i + 1, len(current)):
                    if current[j] == s2[i]:
                        # Swap characters at positions i and j
                        swapped = list(current)
                        swapped[i], swapped[j] = swapped[j], swapped[i]
                        swapped_str = ''.join(swapped)
                        
                        if swapped_str == s2:
                            return swaps + 1
                        
                        if swapped_str not in visited:
                            visited.add(swapped_str)
                            queue.append(swapped_str)
            
            # Increment the number of swaps after processing all nodes at the current level
            swaps += 1
        
        # This point should never be reached as s1 and s2 are anagrams
        return -1

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