Solution Approach

The solution consists of two main parts: preprocessing palindromes and checking for super-palindromes.

Step 1: Generate All Palindromes (Preprocessing)

We iterate through numbers from 1 to 10^5 and generate palindromes:

for i in range(1, 10**5 + 1):  s = str(i)  t1 = s[::-1] # Reverse of the entire string  t2 = s[:-1][::-1] # Reverse of string without last character  ps.append(int(s + t1)) # Even-length palindrome  ps.append(int(s + t2)) # Odd-length palindrome

For example, when i = 123:

• s + t1 creates "123" + "321" = "123321" (even-length)
• s + t2 creates "123" + "21" = "12321" (odd-length)

This generates all palindromes up to approximately 10^10, which when squared can reach up to 10^20, covering our required range.

Step 2: Check if a Number is a Palindrome

The helper function efficiently checks palindrome without string conversion:

def is_palindrome(x: int) -> bool:  y, t = 0, x  while t:  y = y * 10 + t % 10  t //= 10  return x == y

This function reverses the number mathematically:

• Extract the last digit using t % 10
• Build the reversed number by multiplying y by 10 and adding the digit
• Remove the last digit from t using integer division

Step 3: Count Super-Palindromes

l, r = int(left), int(right) return sum(l <= x <= r and is_palindrome(x) for x in map(lambda x: x * x, ps))

For each palindrome p in our precomputed list ps:

1. Calculate x = p² using map(lambda x: x * x, ps)
2. Check if x is within the range [l, r]
3. Check if x itself is a palindrome using is_palindrome(x)
4. Count all numbers satisfying both conditions

Time Complexity: O(M^(1/4) × log M) where M is the upper bound (10^18). We generate O(M^(1/4)) palindromes, and for each palindrome's square, we check if it's a palindrome in O(log M) time.

Space Complexity: O(M^(1/4)) to store all the generated palindromes in the list ps.

Example Walkthrough

Let's walk through finding super-palindromes in the range [4, 1000].

Step 1: Generate palindromes

Starting with small values of i:

When i = 1:

• Even-length: "1" + "1" = "11" → palindrome = 11
• Odd-length: "1" + "" = "1" → palindrome = 1

When i = 2:

• Even-length: "2" + "2" = "22" → palindrome = 22
• Odd-length: "2" + "" = "2" → palindrome = 2

When i = 3:

• Even-length: "3" + "3" = "33" → palindrome = 33
• Odd-length: "3" + "" = "3" → palindrome = 3

When i = 10:

• Even-length: "10" + "01" = "1001" → palindrome = 1001
• Odd-length: "10" + "0" = "100" → palindrome = 100 (not a palindrome, but we still include it)

When i = 11:

• Even-length: "11" + "11" = "1111" → palindrome = 1111
• Odd-length: "11" + "1" = "111" → palindrome = 111

We continue this process up to i = 10^5, building our list ps = [1, 1, 11, 2, 22, 2, 33, 3, ...]

Step 2: Check each palindrome's square

For each palindrome in ps, we square it and check:

• 1² = 1: Is 1 in [4, 1000]? No → skip
• 2² = 4: Is 4 in [4, 1000]? Yes → Is 4 a palindrome? Yes → Count it!
• 3² = 9: Is 9 in [4, 1000]? Yes → Is 9 a palindrome? Yes → Count it!
• 11² = 121: Is 121 in [4, 1000]? Yes → Is 121 a palindrome? Yes → Count it!
• 22² = 484: Is 484 in [4, 1000]? Yes → Is 484 a palindrome? Yes → Count it!
• 33² = 1089: Is 1089 in [4, 1000]? No → skip
• 100² = 10000: Is 10000 in [4, 1000]? No → skip
• 111² = 12321: Is 12321 in [4, 1000]? No → skip

Step 3: Verify palindrome check for 484

Using the is_palindrome function on 484:

• Initialize: y = 0, t = 484
• Iteration 1: y = 0 * 10 + 4 = 4, t = 48
• Iteration 2: y = 4 * 10 + 8 = 48, t = 4
• Iteration 3: y = 48 * 10 + 4 = 484, t = 0
• Compare: 484 == 484 → True, it's a palindrome!

Result: In the range [4, 1000], we found 4 super-palindromes: 4, 9, 121, 484

Time and Space Complexity

Time Complexity: O(n) where n = 10^5

The preprocessing step iterates through numbers from 1 to 10^5, and for each number:

• String conversion takes O(log i) time
• String reversal takes O(log i) time
• Creating palindromes takes O(log i) time
• Total preprocessing: O(n * log n)

For the main function:

• Converting left and right strings to integers: O(1) (assuming fixed-length input)
• Iterating through all palindromes in ps (which has 2 * 10^5 elements): O(n)
• For each palindrome, computing x * x: O(1)
• Checking if x is in range: O(1)
• is_palindrome function: O(log x) where x can be up to (10^10)^2 = 10^20, so O(20) = O(1)
• Total main function: O(n)

Overall time complexity: O(n * log n) for preprocessing + O(n) for query = O(n * log n)

Since preprocessing happens once and n = 10^5 is constant, this simplifies to O(1) amortized per query.

Space Complexity: O(n) where n = 10^5

• The ps list stores 2 * 10^5 palindrome numbers: O(n)
• The is_palindrome function uses O(1) auxiliary space
• Other variables use O(1) space

Total space complexity: O(n) = O(10^5) = O(1) since it's a fixed constant.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Squaring Large Palindromes

The most critical pitfall in this solution is potential integer overflow. When generating palindromes up to 10^5, the resulting palindromes can be as large as 10^10. When squared, these values can reach 10^20, which exceeds the maximum value for standard 32-bit integers and even approaches the limits of 64-bit integers in some languages.

Problem Example:

# A palindrome like 9999999999 (10 digits) # When squared: 9999999999^2 = 99999999980000000001 (20 digits) # This could cause overflow in languages with fixed integer sizes

Solution:

• In Python, this isn't an issue since integers have arbitrary precision
• In other languages like Java or C++, use long long or BigInteger
• Add boundary checks before squaring to prevent overflow

2. Inefficient Palindrome Generation Bounds

Generating palindromes up to 10^5 creates many unnecessary palindromes whose squares will far exceed the typical input range. This wastes memory and computation time.

Problem Example:

# If right = "1000000000" (10^9), we only need palindromes up to sqrt(10^9) ≈ 31623 # But we're generating palindromes up to 10^10, most of which are unnecessary

Solution:

def superpalindromesInRange(self, left: str, right: str) -> int:  right_bound = int(right)  # Calculate the maximum palindrome we need to check  max_palindrome_needed = int(right_bound ** 0.5) + 1   # Generate palindromes only up to what's needed  palindrome_roots = []  limit = min(10**5, max_palindrome_needed)  for num in range(1, limit + 1):  # ... rest of palindrome generation

3. Duplicate Palindromes in the List

The current generation method can create duplicate palindromes, especially for small numbers.

Problem Example:

# When num = 1: # odd_palindrome: "1" + "1" = "11" # even_palindrome: "1" + "" = "1" # When num = 11: # odd_palindrome: "11" + "11" = "1111" # even_palindrome: "11" + "1" = "111" # But "111" might be generated again from num = 111

Solution:

# Use a set to store palindromes and avoid duplicates palindrome_roots = set() for num in range(1, 10**5 + 1):  num_str = str(num)  palindrome_roots.add(int(num_str + num_str[::-1]))  palindrome_roots.add(int(num_str + num_str[:-1][::-1])) palindrome_roots = sorted(list(palindrome_roots))

4. Edge Case: Single-Digit Numbers

The palindrome generation might miss single-digit numbers (1-9), which are palindromes themselves and some of which have palindromic squares.

Problem Example:

# Numbers 1, 4, 9 are super-palindromes (1^2=1, 2^2=4, 3^2=9) # The current generation starts from num=1 but creates "11" and "1" # It might miss handling these edge cases properly

Solution:

# Explicitly include single-digit palindromes palindrome_roots = list(range(1, 10)) # Start with 1-9 for num in range(1, 10**5 + 1):  # ... continue with regular generation

5. String Slicing Index Error

When creating even-length palindromes with num_str[:-1][::-1], if num_str has only one character, num_str[:-1] becomes an empty string.

Problem Example:

# When num = 1 through 9: # num_str = "1", num_str[:-1] = "" (empty string) # This creates the same palindrome twice: "1" + "" = "1"

Solution:

for num in range(1, 10**5 + 1):  num_str = str(num)  palindrome_roots.append(int(num_str + num_str[::-1]))  if len(num_str) > 1: # Only create even-length if meaningful  palindrome_roots.append(int(num_str + num_str[:-1][::-1]))