<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, t1: str, t2: str) -> int:
        @cache
        def solve(i,j):
            if i==0 and j==0:
                return 1 if t1[i]==t2[j] else 0
            if i<0 or j<0 :
                return 0
            r1=0
            r2 = 0
            r3 = 0
            if t1[i]==t2[j]:
                r1 = 1+ solve(i-1,j-1)
            if t1[i]!=t2[j]:
                r2 = solve(i-1,j)
                r3 = solve(i,j-1)
            return max(r1,r2,r3)
        if(t1==t2):
            return 0
        
        ans = solve(len(t1)-1 , len(t2)-1)
        
        return len(t1)- ans + len(t2) - ans
        
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