<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:
        dp = [[float("inf")] * (len(word2) + 1) for i in range(len(word1) + 1)]

        for r in range(len(word1)+1):
            dp[r][len(word2)] = len(word1) - r

        for c in range(len(word2)+1):
            dp[len(word1)][c] = len(word2) - c

        for r in range(len(word1) - 1, -1, -1):
            for c in range(len(word2) - 1, -1, -1):
                if word1[r] == word2[c]:
                    dp[r][c] = dp[r + 1][c + 1]
                else:
                    dp[r][c] = 1 + min(dp[r + 1][c], dp[r][c + 1])
        print(dp)
        return dp[0][0]

        
22428