# Solving Leetcode Interviews in Seconds with AI: Largest Palindrome Product


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "479" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337.   Example 1:  Input: n = 2 Output: 987 Explanation: 99 x 91 = 9009, 9009 % 1337 = 987  Example 2:  Input: n = 1 Output: 9    Constraints:  1 <= n <= 8  

	# Explanation
	Here's a solution to find the largest palindromic integer that can be represented as the product of two n-digit integers, modulo 1337.

*   **High-level approach:** Construct palindromes in descending order and check if each palindrome can be factored into the product of two n-digit numbers. Start with the largest possible n-digit numbers to efficiently find the largest palindromic product. The key is to build the palindrome from the first half, then check divisibility.

*   **Complexity:** Time Complexity: O(10<sup>n</sup>), Storage Complexity: O(1)

	
	# Code
	```python
	def largestPalindromeProduct(n: int) -> int:
    """
    Finds the largest palindromic integer that can be represented as the
    product of two n-digit integers, modulo 1337.

    Args:
        n: The number of digits.

    Returns:
        The largest palindromic product modulo 1337.
    """

    if n == 1:
        return 9

    upper_bound = 10**n - 1
    lower_bound = 10**(n - 1)

    for i in range(upper_bound, lower_bound -1, -1):
        palindrome = int(str(i) + str(i)[::-1])

        for j in range(upper_bound, int(palindrome**0.5) -1, -1):
            if palindrome % j == 0:
                other_factor = palindrome // j
                if (lower_bound <= other_factor <= upper_bound):
                    return palindrome % 1337

    return 0 # Should not reach here given the constraints.
	```