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

        table = [[0 for _ in range(n+1)] for _ in range(m+1)]

        # initialization
        for i in range(m+1):
            table[i][n] = m-i

        for j in range(n+1):
            table[m][j] = n-j

        if word1[m-1] == word2[n-1]:
            table[m][n] = 0
        else:
            table[m][n] = 1

        for i in range(m)[::-1]:
            for j in range(n)[::-1]:
                if word1[i] == word2[j]:
                    table[i][j] = table[i+1][j+1] 
                else:
                    table[i][j] = 1 + min(table[i+1][j], table[i][j+1])

        print(table)

        return table[0][0]

        """
        memo = {}
        def memoization(i, j):

            if (i,j) in memo:
                return memo[(i,j)]
            elif i == m and j < n :
                memo[(i,j)] = 1 + memoization(i, j+1)
            elif i < m and j == n:
                memo[(i,j)] = 1 + memoization(i+1, j)
            elif i == m and j == n:
                memo[(i,j)] = 0
            else:
                if word1[i] == word2[j]:
                    memo[(i,j)] = memoization(i+1, j+1)
                else:
                    memo[(i,j)] = 1 + min(memoization(i+1, j), memoization(i, j+1))

            return memo[(i,j)]
        return memoization(0, 0)
        """
22201