<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:
        
        p, q = len(word1), len(word2)

        #   "" e a t
        # "" 0 1 2 3
        #  s 1 2 3 4
        #  e 2 1 2 3
        #  a 3 2 1 2

        dp = [[float('inf') for _ in range(q + 1)] for _ in range(p + 1)]


        for i in range(p + 1):
            for j in range(q + 1):
                if i == 0 and j == 0:
                    dp[i][j] = 0
                elif i == 0:
                    dp[i][j] = dp[i][j - 1] + 1
                elif j == 0:
                    dp[i][j] = dp[i - 1][j] + 1
                else:
                    if word1[i - 1] != word2[j - 1]:
                        dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
                    else:
                        dp[i][j] = dp[i - 1][j - 1]
        # print(dp)
        return dp[p][q]


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