<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of operations required to convert <code>word1</code> to <code>word2</code></em>.</p>

<p>You have the following three operations permitted on a word:</p>

<ul>
	<li>Insert a character</li>
	<li>Delete a character</li>
	<li>Replace a character</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;horse&quot;, word2 = &quot;ros&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
horse -&gt; rorse (replace &#39;h&#39; with &#39;r&#39;)
rorse -&gt; rose (remove &#39;r&#39;)
rose -&gt; ros (remove &#39;e&#39;)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> word1 = &quot;intention&quot;, word2 = &quot;execution&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
intention -&gt; inention (remove &#39;t&#39;)
inention -&gt; enention (replace &#39;i&#39; with &#39;e&#39;)
enention -&gt; exention (replace &#39;n&#39; with &#39;x&#39;)
exention -&gt; exection (replace &#39;n&#39; with &#39;c&#39;)
exection -&gt; execution (insert &#39;u&#39;)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
</ul>

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:

        # if word1 == "":
        #     return len(word2)
        # elif word2 == "":
        #     return len(word1)
        # else:
        #     loseIt = 1 + min(minDistance(word1, word2[1:]), minDistance(word1[1:], word2))
        #     useIt = word1[0] != word2[0] + minDistance(word1[1:], word2[1:])

        #     return min(loseIt, useIt)    

        # how to do either a bfs or dfs
        
        if word1 == word2:
            return 0
        
        if len(word1) == 0 or len(word2) == 0:
            return max(len(word1), len(word2))

        dq = deque()
        dq.append((0, 0))
        visited = set()
        num = 0
        while len(dq) > 0:
            for _ in range(len(dq)):
                i, j = dq.popleft()
                
                if (i,j) in visited:
                    continue
                
                visited.add((i,j))

                while i < len(word1) and j < len(word2) and word1[i] == word2[j]:
                    i += 1
                    j += 1
                
                if i == len(word1) and j == len(word2):
                    return num

                dq.append((i, j + 1))
                dq.append((i + 1, j))
                dq.append((i + 1, j + 1))

            num += 1



        



        



             


        
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