<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)

        # # dp : O(m.n), O(m.n)
        # @cache
        # def dp(i, j):
        #     if i == m or j == n:
        #         return max(m-i, n-j)
        #     if word1[i] == word2[j]:
        #         return dp(i+1, j+1)
        #     return 1 + min(
        #         dp(i, j+1),
        #         dp(i+1, j),
        #         dp(i+1, j+1)
        #     )
        
        # return dp(0, 0)


        # dp - iterative : O(m.n), O(min(m, n))
        if m > n:
            m, n = n, m
            word1, word2 = word2, word1
        
        dp1 = [0] * (m+1)
        dp2 = [m-i for i in range(m+1)]

        for j in reversed(range(n)):
            dp1[-1] = n-j
            for i in reversed(range(m)):
                if word1[i] == word2[j]:
                    dp1[i] = dp2[i+1]
                else:
                    dp1[i] = 1 + min(
                        dp2[i],
                        dp1[i+1],
                        dp2[i+1]
                    )
            dp1, dp2 = dp2, dp1
        return dp2[0]
16637