<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:
    unordered_map<int,int> dp;
    int recurse(int i, int j, string &s, string &t) {
        if(i > s.size() or j > t.size()) return 1e9;
        if(i == s.size() && j == t.size()) return 0;
        if(dp.find(i*1000+j) != dp.end()) return dp[i*1000+j];

        int res = INT_MAX;
        if(s[i] == t[j]) res = recurse(i+1, j+1, s, t);
        res = min(res, 1 + recurse(i+1, j, s, t));
        res = min(res, 1 + recurse(i, j+1, s, t));
        return dp[i*1000+j] = res;
    }
    int minDistance(string word1, string word2) {
        return recurse(0, 0, word1, word2);
    }
};
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