Introduction

In this blog post, we will explore how to solve the LeetCode problem "906" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Let's say a positive integer is a super-palindrome if it is a palindrome, and it is also the square of a palindrome. Given two positive integers left and right represented as strings, return the number of super-palindromes integers in the inclusive range [left, right]. Example 1: Input: left = "4", right = "1000" Output: 4 Explanation: 4, 9, 121, and 484 are superpalindromes. Note that 676 is not a superpalindrome: 26 * 26 = 676, but 26 is not a palindrome. Example 2: Input: left = "1", right = "2" Output: 1 Constraints: 1 <= left.length, right.length <= 18 left and right consist of only digits. left and right cannot have leading zeros. left and right represent integers in the range [1, 1018 - 1]. left is less than or equal to right.

Explanation

• Palindrome Generation: Generate palindromes efficiently, as the square root of a super-palindrome must also be a palindrome. Construct palindromes up to a certain length.
  • Squaring and Verification: Square the generated palindromes and check if the result is within the given range and is also a palindrome.
  • Optimization: Iterate intelligently to minimize the number of palindrome candidates to square and check, avoiding unnecessary computations.

• Runtime Complexity: O(1), effectively constant due to the constraint on the input range. The number of palindromes generated is small. Storage Complexity: O(1).

Code

    def superpalindromesInRange(left, right):
    """
    Finds the number of super-palindromes in the inclusive range [left, right].

    Args:
        left (str): The left bound of the range.
        right (str): The right bound of the range.

    Returns:
        int: The number of super-palindromes in the range.
    """

    def is_palindrome(s):
        return s == s[::-1]

    left_num = int(left)
    right_num = int(right)
    count = 0

    # Function to generate palindromes
    def generate_palindromes():
        palindromes = []
        # Generate odd length palindromes
        for i in range(1, 100000):
            s = str(i)
            pal = s + s[-2::-1]
            palindromes.append(pal)

        # Generate even length palindromes
        for i in range(1, 100000):
            s = str(i)
            pal = s + s[::-1]
            palindromes.append(pal)

        return palindromes

    palindromes = generate_palindromes()

    for pal_str in palindromes:
        pal_num = int(pal_str)
        super_pal = pal_num * pal_num

        if super_pal >= left_num and super_pal <= right_num and is_palindrome(str(super_pal)):
            count += 1

    return count