<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_v1(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        memo = [[-1 for _ in range(n)] for _ in range(m)]

        def dfs(i: int, j: int):
            if i >= m or j >= n:
                return 0
            if memo[i][j] != -1:
                return memo[i][j]

            memo[i][j] = 0
            if text1[i] == text2[j]:
                memo[i][j] = max(memo[i][j], 1 + dfs(i + 1, j + 1))
            memo[i][j] = max(memo[i][j], dfs(i + 1, j), dfs(i, j + 1))
            return memo[i][j]

        return dfs(0, 0)

    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        memo = [[0 for _ in range(n + 1)] for _ in range(m + 1)]

        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if text1[i] == text2[j]:
                    memo[i][j] = max(memo[i][j], 1 + memo[i + 1][j + 1])
                memo[i][j] = max(memo[i][j], memo[i + 1][j], memo[i][j + 1])

        return memo[0][0]


# s = Solution()
# print(s.longestCommonSubsequence(text1="abc", text2="def"))
# print(s.longestCommonSubsequence(text1="abc", text2="abc"))
# print(s.longestCommonSubsequence(text1="abcde", text2="ace"))

40103