Introduction

In this blog post, we will explore how to solve the LeetCode problem "866" 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

Given an integer n, return the smallest prime palindrome greater than or equal to n. An integer is prime if it has exactly two divisors: 1 and itself. Note that 1 is not a prime number. For example, 2, 3, 5, 7, 11, and 13 are all primes. An integer is a palindrome if it reads the same from left to right as it does from right to left. For example, 101 and 12321 are palindromes. The test cases are generated so that the answer always exists and is in the range [2, 2 * 108]. Example 1: Input: n = 6 Output: 7 Example 2: Input: n = 8 Output: 11 Example 3: Input: n = 13 Output: 101 Constraints: 1 <= n <= 108

Explanation

Here's a breakdown of the approach and the Python code:

• Approach:

  • Generate palindromes starting from the smallest possible palindrome greater than or equal to the input n.
  • Check each generated palindrome for primality.
  • Return the first palindrome that is both prime and greater than or equal to n. We optimize by generating only odd length palindromes for larger numbers, and handling the special case of single digit numbers.

• Complexity:

  • Runtime Complexity: O(sqrt(N) * log(N)), where N is the result. The log(N) factor arises because of the palindrome generation, and the sqrt(N) is due to primality testing.
  • Storage Complexity: O(1)

Code

    import math

def is_prime(n):
    if n < 2:
        return False
    for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0:
            return False
    return True

def generate_palindrome(length):
    start = 10 ** ((length - 1) // 2)
    end = 10 ** ((length + 1) // 2)

    for i in range(start, end):
        s = str(i)
        palindrome = s + s[:(length // 2)][::-1]
        yield int(palindrome)

def prime_palindrome(n):
    if n <= 2:
        return 2
    if n <= 3:
        return 3
    if n <= 5:
        return 5
    if n <= 7:
        return 7
    if n <= 11:
        return 11

    length = len(str(n))
    if length % 2 == 0:
        length += 1

    while True:
        for palindrome in generate_palindrome(length):
            if palindrome >= n and is_prime(palindrome):
                return palindrome
        length += 2