<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[510][510];
    int rec(int i , int j , string s, string t){
        if(i < 0 && j < 0)return 0;
        if(i < 0 && j >= 0)return j+1;
        if(i >= 0 && j < 0)return i+1;
        if(dp[i][j] != -1)return dp[i][j];
        int ans = 0;

        if(s[i] == t[j])ans =rec(i-1 , j-1 , s , t);
        else ans = min(1 + rec(i-1 , j , s , t) , 1 + rec(i , j-1 , s , t));

        return dp[i][j] = ans;
    }
    int minDistance(string word1, string word2) {
        memset(dp , -1 , sizeof(dp));
        return rec(word1.size()-1 , word2.size()-1 , word1 , word2);
    }
};
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