<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:
        def solve(text):
            mp = {}
            for i in range(len(text)):
                half = text[i]
                k = i
                j = i + 2
                while j <= len(text):
                    # print(half,text[i:j],i,j)
                    if half + half == text[i:j]:
                        mp[half] = 1
                    j+=2
                    k+=1
                    if j <= len(text):
                        half+=text[k]
            if len(text)%2 == 0:
                if text[:len(text)//2] + text[:len(text)//2] == text:
                    mp[text[:len(text)//2]] = 1
            print(mp)
            if "" in mp:
                return len(mp)-1
            else:
                return len(mp)
            # return 2

        return solve(text)
18100