<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:
    seen = set()

    for k in range(1, len(text) // 2 + 1):  # the target length
      same = 0
      l = 0
      for r in range(k, len(text)):
        if text[l] == text[r]:
          same += 1
        else:
          same = 0
        if same == k:
          seen.add(text[l - k + 1:l + 1])
          # Move the window thus leaving a letter behind, so we need to
          # decrease the counter.
          same -= 1
        l += 1

    return len(seen)
17000