<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:
        if s1 == s2:
            return 0
        
        # Initial state
        queue = deque([(s1, 0)])  # (current string, number of swaps)
        visited = set()  # To track visited states
        visited.add(s1)
        
        while queue:
            current, swaps = queue.popleft()
            
            # Generate all possible swaps
            for i in range(len(current)):
                if current[i] != s2[i]:
                    # Find the first position where current differs from s2
                    j = i + 1
                    while j < len(current) and (current[j] == s2[j] or current[i] != s2[j]):
                        j += 1
                    
                    if j < len(current):
                        # Swap the characters at positions i and j
                        next_state = list(current)
                        next_state[i], next_state[j] = next_state[j], next_state[i]
                        next_state = ''.join(next_state)
                        if next_state == s2:
                            return swaps + 1
                        
                        if next_state not in visited:
                            visited.add(next_state)
                            queue.append((next_state, swaps + 1))
        
        return -1  # Return -1 if no solution is found, though there should always be one in valid inputs.

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