<p>Given two strings <code>word1</code> and <code>word2</code>, return <em>the minimum number of <strong>steps</strong> required to make</em> <code>word1</code> <em>and</em> <code>word2</code> <em>the same</em>.</p>

<p>In one <strong>step</strong>, you can delete exactly one character in either string.</p>

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

<pre>
<strong>Input:</strong> word1 = &quot;sea&quot;, word2 = &quot;eat&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need one step to make &quot;sea&quot; to &quot;ea&quot; and another step to make &quot;eat&quot; to &quot;ea&quot;.
</pre>

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

<pre>
<strong>Input:</strong> word1 = &quot;leetcode&quot;, word2 = &quot;etco&quot;
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code> and <code>word2</code> consist of only lowercase English letters.</li>
</ul>

import numpy as np
class Solution:
    def minDistance(self, word1: str, word2: str) -> int:

        res = np.zeros((len(word1), len(word2)))

        for i in range(len(word1)):
            if word1[i] != word2[0]:
                res[i,0] = i+2
            else:
                for k in range(i, len(word1)):
                    res[k,0] = k
                break
        
        for j in range(len(word2)):
            if word2[j] != word1[0]:
                res[0,j] = j+2
            else:
                for k in range(j, len(word2)):
                    res[0,k] = k
                break
        
        D = dict() # word1
        D[word1[0]] = 0
        for i in range(1, len(word1)):
            D[word1[i]] = i
            for j in range(1, len(word2)):
                boo = word2[j] in D
                if boo and D[word2[j]] > 0:
                    res[i,j] = min(1+res[i,j-1], i-D[word2[j]]+res[D[word2[j]]-1,j-1])
                elif not boo:
                    res[i,j] = 1+res[i,j-1]
                else:
                    res[i,j] = min(1+res[i,j-1], i+j)

        return int(res[-1,-1])
        
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