<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>

from collections import defaultdict
class Solution:
    def print_dp(self, dp, n, m):
        for i in range(n):
            ret = [str(dp[(i, x)]) for x in range(m)]
            print(','.join(ret))
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = defaultdict(lambda: 0)
        row, col = 0, 0

        while row < len(text1) and col < len(text2):
            for r in range(row, len(text1)):
                if text1[r] == text2[col]:
                    dp[(r, col)] += 1 + dp[(r - 1, col - 1)]
                else:
                    dp[(r, col)] = max(dp[(r - 1, col)], dp[(r, col - 1)])
            for c in range(col + 1, len(text2)):
                if text1[row] == text2[c]:
                    dp[(row, c)] += 1 + dp[(row - 1, c - 1)]
                else:
                    dp[(row, c)] = max(dp[(row - 1, c)], dp[(row, c - 1)])
            row += 1
            col += 1

        #self.print_dp(dp, len(text1), len(text2))
        return dp[(len(text1) - 1, len(text2) - 1)]


        
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