<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>

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

        previous = list(range(1,n+1))

        current = [-1] * n
        for i in range(0, len(word1)):
            if word1[i] == word2[0]:
                current[0] = i
            else:
                current[0] = min(i + 1, previous[0]) + 1
            for j in range(1, len(word2)):
                if word1[i] == word2[j]:
                    current[j] = previous[j - 1]
                else:
                    current[j] = min(current[j - 1], previous[j]) + 1
            previous = current.copy()
            print(f'{i}, {current}')
        return current[-1]


        
16741