<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:

      # @cache
      # def dfs(i, j):
      #   if i == len(word1) and j == len(word2):
      #     return 0
      #   elif i == len(word1):
      #     return len(word2) - j
      #   elif j == len(word2):
      #     return len(word1) - i
      #   else:
      #     if word1[i] == word2[j]:
      #       return dfs(i + 1, j + 1)
      #     return 1 + min(dfs(i + 1, j), dfs(i, j + 1))
      
      # return dfs(0, 0)

      n = len(word1)
      m = len(word2)

      table = [[float("inf") for _ in range(m  + 1)] for _ in range(n + 1)]

      table[0][0] = 0

      for row in range(1, n + 1):
        table[row][0] = row
      for col in range(1, m + 1):
        table[0][col] = col
      
      for row in range(1, n + 1):
        for col in range(1, m + 1):
          if word1[row - 1] == word2[col - 1]:
            table[row][col] = table[row - 1][col - 1]
          else:
            table[row][col] = 1 + min(table[row - 1][col], table[row][col - 1])
      print(table) 
      return table[n][m]


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