<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:
        """
        i, j
        dp(i, j) the longest substr until i-th str
        base: i == len(text1) return 0
        
        """
        # time O(mn), space O(mn)
        if len(text1) > len(text2):
            text1, text2 = text2, text1
        
        memo = {}
        def dp(i,j):
            if (i,j) in memo:
                return memo[(i,j)]
            if i == len(text1) or j == len(text2):
                return 0
            # select
            p = j
            while p < len(text2) and text2[p] != text1[i]:
                p += 1
            if p < len(text2):
                op1 = 1 + dp(i+1,p+1)
            else:
                op1 = 0

            # not select
            op2 = dp(i+1,j)
            memo[(i,j)] = max(op1, op2)
            return memo[(i,j)]

        return dp(0,0)

            # if text1[i] == text2[j]:
            #     memo[(i,j)] = max(
            #         1 + dp(i+1,j+1),
            #         dp(i,j+1),
            #         dp(i+1,j)
            #     )
            # else:
            #     memo[(i,j)] = max(
            #         dp(i,j+1),
            #         dp(i+1,j)

            #     )
        #     return memo[(i,j)]
        # return dp(0,0)


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