<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, t1: str, t2: str) -> int:
        lt1 = len(t1)
        lt2 = len(t2)
        dp = [[0 for i in range(lt1+1)] for i in range(lt2+1)]
        q = deque()
        q.append((lt2-1,lt1-1))
        seen = set()
        res = 0
        while q:
            r,c = q.popleft()
            if (r,c) not in seen and 0<=r<lt2 and 0<=c<lt1:
                
                
                if t2[r] == t1[c]:
                    print(t2[r],t1[c])
                    dp[r][c] = 1 + dp[r+1][c+1]
                else:
                    dp[r][c] = max(dp[r+1][c] ,  dp[r][c+1])

                res = max(res, dp[r][c])
                seen.add((r,c))
                q.append((r-1,c))
                q.append((r,c-1))
        print(dp)
        return res
194558