<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') for _ in range(len(word2))] for _ in range(len(word1))]

        def helper(i, j):
            if i >= len(word1) or j >= len(word2):
                return 0
            if dp[i][j] != float('-inf'):
                return dp[i][j]

            ret = float('-inf')
            if word1[i] == word2[j]:
                dp[i][j] =  1 + helper(i+1, j+1)
                return dp[i][j]
            ret = max(ret, helper(i+1, j))
            ret = max(ret, helper(i, j+1))
            dp[i][j] = ret
            return ret
        helper(0, 0)
        print(dp)
        return len(word1) + len(word2) - 2*dp[0][0]
24703