<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:
        # recursive code

        # try to understand the working of recursive function and
        # and have believe on it
        memo = {}
        
        def dp(i, j):
            # Base cases:
            if i == len(word1):
                return len(word2) - j  # Remaining characters in word2
            if j == len(word2):
                return len(word1) - i  # Remaining characters in word1

            # If result is already calculated
            if (i, j) in memo:
                return memo[(i, j)]

            # If characters match, no deletion needed, move to the next character in both strings
            if word1[i] == word2[j]:
                memo[(i, j)] = dp(i + 1, j + 1)
            else:
                
                delete_word1 = dp(i + 1, j)  # Delete from word1
                delete_word2 = dp(i, j + 1)  # Delete from word2
                memo[(i, j)] = 1 + min(delete_word1, delete_word2)

            return memo[(i, j)]
            
        return dp(0, 0)




        # top down appraoch
        def lcs(s1, s2):
            m,n = len(s1), len(s2)

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

            for i in range(1, m+1):
                for j in range(1, n+1):
                    if s1[i-1] == s2[j-1]:
                        dp[i][j] = dp[i-1][j-1] + 1
                    else:
                        dp[i][j] = max(dp[i-1][j], dp[i][j-1])
            return dp[-1][-1]

        common = lcs(word1, word2)

        return len(word1) + len(word2) -2*common
28571