<p>Return the number of <strong>distinct</strong> non-empty substrings of <code>text</code>&nbsp;that can be written as the concatenation of some string with itself (i.e. it can be written as <code>a + a</code>&nbsp;where <code>a</code> is some string).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;abcabcabc&quot;
<strong>Output:</strong> 3
<b>Explanation: </b>The 3 substrings are &quot;abcabc&quot;, &quot;bcabca&quot; and &quot;cabcab&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;leetcodeleetcode&quot;
<strong>Output:</strong> 2
<b>Explanation: </b>The 2 substrings are &quot;ee&quot; and &quot;leetcodeleetcode&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text.length &lt;= 2000</code></li>
	<li><code>text</code>&nbsp;has only lowercase English letters.</li>
</ul>

class Solution:
    def distinctEchoSubstrings(self, text: str) -> int:
        n = len(text)
        seen = set()
        
        # Iterate over all possible even lengths
        for length in range(2, n + 1, 2):
            for i in range(n - length + 1):
                mid = i + length // 2
                if text[i:mid] == text[mid:i + length]:
                    seen.add(text[i:i + length])
        
        return len(seen)

# Example usage:
solution = Solution()
print(solution.distinctEchoSubstrings("abcabcabc"))  # Output: 3
print(solution.distinctEchoSubstrings("leetcodeleetcode"))  # Output: 2
17500