<p>Let&#39;s say a positive integer is a <strong>super-palindrome</strong> if it is a palindrome, and it is also the square of a palindrome.</p>

<p>Given two positive integers <code>left</code> and <code>right</code> represented as strings, return <em>the number of <strong>super-palindromes</strong> integers in the inclusive range</em> <code>[left, right]</code>.</p>

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

<pre>
<strong>Input:</strong> left = &quot;4&quot;, right = &quot;1000&quot;
<strong>Output:</strong> 4
<strong>Explanation</strong>: 4, 9, 121, and 484 are superpalindromes.
Note that 676 is not a superpalindrome: 26 * 26 = 676, but 26 is not a palindrome.
</pre>

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

<pre>
<strong>Input:</strong> left = &quot;1&quot;, right = &quot;2&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= left.length, right.length &lt;= 18</code></li>
	<li><code>left</code> and <code>right</code> consist of only digits.</li>
	<li><code>left</code> and <code>right</code> cannot have leading zeros.</li>
	<li><code>left</code> and <code>right</code> represent integers in the range <code>[1, 10<sup>18</sup> - 1]</code>.</li>
	<li><code>left</code> is less than or equal to <code>right</code>.</li>
</ul>

class Solution:
    def is_palindrome(self, x: str) -> bool:
        return x == x[::-1]

    def superpalindromesInRange(self, left: str, right: str) -> int:
        L = int(left)
        R = int(right)
        MAGIC = 100000  # We generate palindrome roots up to MAGIC number
        
        count = 0
        
        # Check odd length palindromes
        for k in range(1, MAGIC):
            s = str(k)
            # Create odd length palindrome (like 12321)
            palindrome = int(s + s[-2::-1])
            square = palindrome * palindrome
            if square > R:
                break
            if square >= L and self.is_palindrome(str(square)):
                count += 1
        
        # Check even length palindromes
        for k in range(1, MAGIC):
            s = str(k)
            # Create even length palindrome (like 123321)
            palindrome = int(s + s[::-1])
            square = palindrome * palindrome
            if square > R:
                break
            if square >= L and self.is_palindrome(str(square)):
                count += 1
        
        return count
16400