<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:
        m, n = len(word1), len(word2)
        arr = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for i in range(m, -1, -1):
            arr[i][-1] = m-i
        for j in range(n, -1, -1):
            arr[-1][j] = n-j
        for i in range(m-1, -1, -1):
            for j in range(n-1, -1, -1):
                if word1[i] == word2[j]:
                    arr[i][j] = arr[i+1][j+1]
                else:
                    arr[i][j] = min(arr[i+1][j+1], arr[i+1][j], arr[i][j+1]) + 1
        print(arr)
        return arr[0][0]

        
        def dfs(i, j):
            if (i, j) in memo:
                return memo[(i, j)]
            if i == len(word1):
                return len(word2)-j
            if j == len(word2):
                return len(word1)-i
            if word1[i] == word2[j]:
                ans = dfs(i+1, j+1)
            else:
                ans = min(dfs(i+1, j+1), dfs(i, j+1), dfs(i+1, j)) + 1
            memo[(i, j)] = ans
            return ans
        memo = dict()
        return dfs(0, 0)
22762