<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:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:

        def recComSeq(si1, si2):
            if dp[si1][si2] != -1:
                return dp[si1][si2]
            if si1 == 0 or si2 == 0:
                return 0
            else:
                # pick
                p = 0
                if text1[si1 - 1 ] ==  text2[si2 - 1]:
                    p = 1 + recComSeq(si1 - 1, si2 - 1)
                    dp[si1][si2] = p
                    return p

                q = recComSeq(si1 - 1, si2)
                r = recComSeq(si1, si2 - 1)
                dp[si1][si2] = max(q, r)
                return max(q, r)

        m = len(text1)
        n = len(text2)
        dp = [[-1]*(n + 1) for i in range(m + 1)]
        for i in range(m+1):
            dp[i][0] = 0
        for j in range(n+1):
            dp[0][j] = 0


        for si1 in range(1, m+1):
            for si2 in range(1, n+1):
                # pick
                p = 0
                if text1[si1 - 1 ] ==  text2[si2 - 1]:
                    p = 1 + dp[si1 - 1][si2 - 1]
                    dp[si1][si2] = p
                else:
                    q = dp[si1 - 1][si2]
                    r = dp[si1][si2 - 1]
                    dp[si1][si2] = max(q, r)
        return dp[m][n]

        
58491