<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:
        m, n = len(text1), len(text2)
        memo = [ [-1] * n for _ in range(m) ]

        def memoization(idx1, idx2):
            # base cases
            if idx1 < 0 or idx2 < 0:
                return 0
            if memo[idx1][idx2] != -1:
                return memo[idx1][idx2]
            
            # if same letters, decrement both else try 2 combos available by decrementing each seperately
            if text1[idx1] == text2[idx2]:
                memo[idx1][idx2] = 1 + memoization(idx1-1, idx2-1)
            else:
                memo[idx1][idx2] = max(memoization(idx1-1, idx2), memoization(idx1, idx2-1))

            return memo[idx1][idx2]
        
        # trying memoization
        # return memoization(m-1, n-1)


        def tabulation():
            dp = [ [-1] * (n+1) for _ in range(m+1) ]

            # base cases
            for i in range(n+1):
                dp[0][i] = 0
            for j in range(m+1):
                dp[j][0] = 0
            print(dp)
            for i in range(1, m+1):
                for j in range(1, n+1):
                    if text1[i-1] == text2[j-1]:
                        dp[i][j] = 1 + dp[i-1][j-1]
                    else: 
                        dp[i][j] = max(dp[i-1][j], dp[i][j-1])
            
            return dp[m][n]
        
        # trying tabulation
        return tabulation()

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