How do you solve Super Palindromes on LeetCode?

Super Palindromes (LeetCode #906) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Super-palindromes must be both palindromic and squares of palindromic numbers.

What is the brute force solution for Super Palindromes?

The brute force approach for Super Palindromes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every number in the range from left to right, determining if it is a palindrome and if its square is also a palindrome. This method is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Super Palindromes?

Super Palindromes uses the following concepts: Math, String, Enumeration. The recommended patterns to study are: Enumeration, Two Pointers.

What are the interview tips for Super Palindromes?

Always clarify the range and constraints before starting. Think about generating numbers instead of checking all possibilities. Practice identifying patterns in problems to speed up your solution process.

What are common mistakes when solving Super Palindromes?

Failing to check both conditions for super-palindromes. Not considering the range limits when generating palindromic numbers.

#906

Super Palindromes

Hard

MathStringEnumerationEnumerationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can generate palindromic numbers directly and only check their squares. This reduces the number of checks significantly and avoids iterating through all numbers in the range.

⚙️

Algorithm

4 steps

1Step 1: Generate palindromic numbers up to the square root of right.
2Step 2: For each palindromic number, compute its square.
3Step 3: Check if the square is a palindrome and falls within the range [left, right].
4Step 4: Count valid super-palindromes.

solution.py18 lines

1def is_palindrome(x): return str(x) == str(x)[::-1]
2
3def super_palindromes(left, right):
4    left, right = int(left), int(right)
5    count = 0
6    for i in range(1, 100000):
7        s = str(i)
8        pal = int(s + s[::-1])  # Even length palindromes
9        pal_sq = pal * pal
10        if pal_sq > right: break
11        if pal_sq >= left and is_palindrome(pal_sq):
12            count += 1
13        pal = int(s + s[-2::-1])  # Odd length palindromes
14        pal_sq = pal * pal
15        if pal_sq > right: break
16        if pal_sq >= left and is_palindrome(pal_sq):
17            count += 1
18    return count

ℹ

Complexity note: This complexity is efficient because we only generate palindromic numbers and check their squares, significantly reducing the number of iterations.

1Super-palindromes must be both palindromic and squares of palindromic numbers.
2Generating palindromic numbers directly is more efficient than checking every number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.