<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>

'''
define dp: dp[i][j] means how many same letters (in order) in common between word1[0:i] and word2[0:j]
if word1[i] ==  word2[j]: dp[i][j] = dp[i-1][j-1] +1
else: dp[i][j] = max(dp[i-1][j], dp[i][j-1])
'''
class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0 for _ in range(len(word2) + 1)] for _ in range(len(word1) + 1)]

        for i in range(0, len(word1)):
            for j in range(0, len(word2)):
                if word1[i] == word2[j]:
                    dp[i+1][j+1] = dp[i][j] +1
                else:
                    dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1])

        return len(word1) + len(word2) - 2*(dp[-1][-1])



        
18106