<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 {
public:
    int dp[505][505];

    int rec(int i,int j, string x, string y){
        int n=x.length(), m=y.length();
        if(i==n || j==m) return max(n-i, m-j);

        if(dp[i][j]!=-1) return dp[i][j];

        int ans=1e9;
        if(x[i]==y[j]) 
            ans= min(ans, rec(i+1,j+1,x,y));
        else
            ans=min({ans, 1+rec(i, j+1, x, y), 1+rec(i+1, j, x, y)});
        // rec(i,j+1)-to insert,  rec(i+1,j)-to delete,  rec(i+1,j+1)-to replace

        return dp[i][j]=ans;
    }

    int minDistance(string word1, string word2) {
        memset(dp,-1,sizeof(dp));
        return rec(0,0,word1,word2);
    }
};
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