<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 superpalindromesInRange(self, left: str, right: str) -> int:
    def nextPalindrome(num: int) -> int:
      s = str(num)
      n = len(s)

      half = s[0:(n + 1) // 2]
      reversedHalf = half[:n // 2][::-1]
      candidate = int(half + reversedHalf)
      if candidate >= num:
        return candidate

      half = str(int(half) + 1)
      reversedHalf = half[:n // 2][::-1]
      return int(half + reversedHalf)

    def isPalindrome(num: int) -> bool:
      s = str(num)
      l = 0
      r = len(s) - 1

      while l < r:
        if s[l] != s[r]:
          return False
        l += 1
        r -= 1

      return True

    ans = 0
    l = int(left)
    r = int(right)
    i = math.isqrt(l)

    while i * i <= r:
      palindrome = nextPalindrome(i)
      squared = palindrome**2
      if squared <= r and isPalindrome(squared):
        ans += 1
      i = palindrome + 1

    return ans
16500