<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>

def solve(i,j,s1,s2,dp):
    if i<0 and j<0:
        return 0
    if i<0 or j<0:
        if i>=0:
            return i+1
        else:
            return j+1
    if dp[i][j]!=-1:
        return dp[i][j]
    if s1[i]==s2[j]:
        ans=solve(i-1,j-1,s1,s2,dp)
    else:
        del1=1+solve(i-1,j,s1,s2,dp)
        del2=1+solve(i,j-1,s1,s2,dp)
        ans=min(del1,del2)
    dp[i][j]=ans
    return ans 
class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        s1,s2=word1,word2
        n,m=len(word1),len(word2)
        dp=[[0 for i in range(m+1)]for i in range(n+1)]
        for i in range(n+1):
            dp[i][0]=i
        for j in range(m+1):
            dp[0][j]=j

        for i in range(1,n+1):
            for j in range(1,m+1):
            
                if s1[i-1]==s2[j-1]:
                    ans=dp[i-1][j-1]
                else:
                    del1=1+dp[i-1][j]
                    del2=1+dp[i][j-1]
                    ans=min(del1,del2)
                dp[i][j]=ans
        return dp[n][m]
        # ans=solve(n-1,m-1,word1,word2,dp)
        # return ans 
20153