<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:
    dp=[]
    def helper(self, text1, i, text2, j):
        if i==len(text1) or j==len(text2):
            return 0
        
        if self.dp[i][j]!=-1:
            return self.dp[i][j]

        ans=0
        if text1[i]==text2[j]:
            ans = 1+self.helper(text1,i+1,text2,j+1) #got one character common
        else:
            op1 = self.helper(text1,i+1,text2,j)
            op2 = self.helper(text1,i,text2,j+1)
            ans = max(op1,op2)
        
        #ans variable will be memoised
        self.dp[i][j] = ans
        return ans
            
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        self.dp = [[-1]*len(text2) for i in range(len(text1))]
        print(self.dp)
        return self.helper(text1, 0, text2, 0)



# text1 = abce
# text2 = bcea

# ans = 3 bce

# text1 = cdefd
# text2 = cefg


# 1 = abcde
# i=0
# 2 = ace
# j=0

# i==j==> 1 + 

# 1 = abcde
# i=1
# 2 = ace
# j=1


# memoisation grid
# 1D array = linear order
# 2d grid = quadratic nature


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