<p>Given two strings <code>text1</code> and <code>text2</code>, return <em>the length of their longest <strong>common subsequence</strong>. </em>If there is no <strong>common subsequence</strong>, return <code>0</code>.</p>

<p>A <strong>subsequence</strong> of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.</p>

<ul>
	<li>For example, <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;abcde&quot;</code>.</li>
</ul>

<p>A <strong>common subsequence</strong> of two strings is a subsequence that is common to both strings.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abcde&quot;, text2 = &quot;ace&quot; 
<strong>Output:</strong> 3  
<strong>Explanation:</strong> The longest common subsequence is &quot;ace&quot; and its length is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;abc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest common subsequence is &quot;abc&quot; and its length is 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;def&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no such common subsequence, so the result is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text1.length, text2.length &lt;= 1000</code></li>
	<li><code>text1</code> and <code>text2</code> consist of only lowercase English characters.</li>
</ul>

class Solution {
    string text1_, text2_;
    int dp[1000][1000];
    bool seen[1000][1000];

    int lcs(int i, int j) {
        if (i < 0 || j < 0)
            return 0;
        if (seen[i][j])
            return dp[i][j];
        
        seen[i][j] = true;
        // cout << i << " " << j << endl;
        if (text1_[i] == text2_[j])
            dp[i][j] = 1 + lcs(i - 1, j - 1);
        else
            dp[i][j] = max(lcs(i - 1, j), lcs(i, j - 1));
        // cout << i << " " << j << " " << dp[i][j] << endl;

        return dp[i][j];
    }
public:
    int longestCommonSubsequence(string text1, string text2) {
        text1_ = text1;
        text2_ = text2;

        return lcs(text1_.size() - 1, text2_.size() - 1);
    }
};
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