<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 lcsImpl(string text1, string text2, int idx1, int idx2, vector<vector<int>>& book) {
        if (idx1 == text1.size() || idx2 == text2.size()) {
            return 0;
        }
        if (book[idx1][idx2] != -1) {
            return book[idx1][idx2];
        }
        if (text1[idx1] == text2[idx2]) {
            return book[idx1][idx2] = 1 + lcsImpl(text1, text2, idx1 + 1, idx2 + 1, book);
        } else {
            return book[idx1][idx2] = max(lcsImpl(text1, text2, idx1, idx2 + 1, book), 
                lcsImpl(text1, text2, idx1 + 1, idx2, book));
        }
    }
    int longestCommonSubsequence(string text1, string text2) {
        vector<vector<int>> book(text1.size() + 1, vector<int>(text2.size() + 1, -1));
        return lcsImpl(text1, text2, 0, 0, book);
    }
    int minDistance(string word1, string word2) {
        int lcs = longestCommonSubsequence(word1, word2);
        return word1.size() - lcs + word2.size() - lcs;
    }
};
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