Introduction

In this blog post, we will explore how to solve the LeetCode problem "2081" 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

A k-mirror number is a positive integer without leading zeros that reads the same both forward and backward in base-10 as well as in base-k. For example, 9 is a 2-mirror number. The representation of 9 in base-10 and base-2 are 9 and 1001 respectively, which read the same both forward and backward. On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100, which does not read the same both forward and backward. Given the base k and the number n, return the sum of the n smallest k-mirror numbers. Example 1: Input: k = 2, n = 5 Output: 25 Explanation: The 5 smallest 2-mirror numbers and their representations in base-2 are listed as follows: base-10 base-2 1 1 3 11 5 101 7 111 9 1001 Their sum = 1 + 3 + 5 + 7 + 9 = 25. Example 2: Input: k = 3, n = 7 Output: 499 Explanation: The 7 smallest 3-mirror numbers are and their representations in base-3 are listed as follows: base-10 base-3 1 1 2 2 4 11 8 22 121 11111 151 12121 212 21212 Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499. Example 3: Input: k = 7, n = 17 Output: 20379000 Explanation: The 17 smallest 7-mirror numbers are: 1, 2, 3, 4, 5, 6, 8, 121, 171, 242, 292, 16561, 65656, 2137312, 4602064, 6597956, 6958596 Constraints: 2 <= k <= 9 1 <= n <= 30

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Generate Palindromes: The core idea is to efficiently generate palindromic numbers in base-10 and then check if they are also palindromic in base-k. We can construct palindromes by building half of the number and then mirroring it.

• Base-k Conversion and Palindrome Check: We need a function to convert a number to base-k and another function to efficiently check if a given string (representing the number in base-k) is a palindrome.

• Iterative Generation and Summation: We generate palindromes of increasing length until we find 'n' k-mirror numbers. We sum these numbers to obtain the final result.

• Runtime and Storage Complexity:

  • Runtime Complexity: O(n * logk(M)), where M is the largest k-mirror number and logk(M) is the cost of converting M to base k and checking if it is a palindrome. We generate palindromes up to a certain length, but the number of such palindromes scales exponentially with the length, and the base-k conversion and check take logarithmic time.
  • Storage Complexity: O(logk(M)). Dominated by the space required to store the largest k-mirror number in base k.

Code

    def is_k_palindrome(n, k):
    """Checks if a number n is a palindrome in base-k."""
    if n < 0:
        return False

    digits = []
    while n > 0:
        digits.append(n % k)
        n //= k

    l, r = 0, len(digits) - 1
    while l < r:
        if digits[l] != digits[r]:
            return False
        l += 1
        r -= 1
    return True


def generate_palindromes():
    """Generates palindromes of increasing length."""
    yield 1  # Special case for 1

    for length in range(1, 11):  # Adjust max length as needed
        # Odd length palindromes
        for i in range(1, 10):  # First digit cannot be 0
            num_str = str(i)
            palindrome = int(num_str)
            yield palindrome
            for j in range(1, (length + 1) // 2):
                for middle_digit in range(0, 10):
                    num_str = str(i)
                    middle = ""
                    temp = num_str
                    for x in range(j):
                        temp = "0" + temp
                    num_str = temp
                    str_num = str(i)


                    middle = str(middle_digit)
                    palindrome_str = str_num + middle
                    for x in range(j):
                        palindrome_str = "0" + palindrome_str

                    palindrome = int(palindrome_str + str(middle_digit) + str(i))


                    if(len(str(palindrome)) == length):
                        yield palindrome

        # Even Length Palindromes
        for i in range(1, 10):
            num_str = str(i)
            palindrome = int(num_str + num_str)
            yield palindrome
            for j in range(1, (length) // 2):
                for middle_digit in range(0, 10):
                    num_str = str(i)
                    middle = ""
                    temp = num_str
                    for x in range(j):
                        temp = "0" + temp
                    num_str = temp
                    str_num = str(i)


                    middle = str(middle_digit)
                    palindrome_str = str_num + middle
                    for x in range(j):
                        palindrome_str = "0" + palindrome_str

                    palindrome = int(palindrome_str + str(middle_digit) + str(i))


                    if(len(str(palindrome)) == length):
                        yield palindrome


def k_mirror(k, n):
    """Calculates the sum of the n smallest k-mirror numbers."""
    count = 0
    total_sum = 0

    for p in generate_palindromes():
        if is_k_palindrome(p, k):
            total_sum += p
            count += 1
            if count == n:
                break

    return total_sum