<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 queue import Queue
from heapq import heapify, heappop, heappush
import numpy as np

def swapPos(s, i, j):
    # Convert the string to a list to allow modifications
    s_list = list(s)
    
    # Swap the characters at indices i and j
    s_list[i], s_list[j] = s_list[j], s_list[i]
    
    # Convert the list back to a string
    v= ''.join(s_list)
    #print(v)
    return v    

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        def heur(curr, target):
            count = 0
            for i in range(len(curr)):
                if curr[i] != target[i]:
                    count+=1
            #print("heur. curr:", curr, "  target:", target, "  res:", count//2)
            r = -2 + count//2
            mult =  (1 - np.exp(-0.3 * (r)))
            #print(r, "=>", mult)
            return r * mult
        if s1==s2: return 0
        visited = set()
        #queue = Queue()
        q = []
        #queue.put((0, s1))
        heappush(q, (0,0,s1))
        visited.add(s1)
        while q:
            heurVal, distance, s = heappop(q)
            
            i=0
            while s[i] == s2[i]:
                i+=1
            for j in range(i+1, len(s)):
                #newNode = None
                if s[j] == s2[j]: # if j is already correct
                    continue
                elif s[j] == s2[i]:  # if this is what we want at i
                    newNode = swapPos(s, i, j)
                    if newNode == s2:
                        return distance+1
                    if not newNode in visited:
                        visited.add(newNode)
                        heappush(q, (distance+1+heur(newNode, s2), distance+1, newNode))
            #badIndices = []
            #for i in range(len(s)):
            #    if s[i] != s2[i]:
            #        badIndices.append(i)
            ##print(distance, s, badIndices)
            ##print(badIndices)
            ##return 
            #for i in range(len(badIndices)):
            #    for j in range(i+1, len(badIndices)):
            #        ind1 = badIndices[i]
            #        ind2 = badIndices[j]
            #        newNode = swapPos(s, ind1, ind2)
            #        if newNode == s2:
            #            return distance+1
            #        if not newNode in visited:
            #            visited.add(newNode)
            #            heappush(q, (distance+1+heur(newNode, s2), distance+1, newNode))
            
        return -1
30000