Introduction

In this blog post, we will explore how to solve the LeetCode problem "3233" 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

You are given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x. A number is called special if it has exactly 2 proper divisors. For example: The number 4 is special because it has proper divisors 1 and 2. The number 6 is not special because it has proper divisors 1, 2, and 3. Return the count of numbers in the range [l, r] that are not special. Example 1: Input: l = 5, r = 7 Output: 3 Explanation: There are no special numbers in the range [5, 7]. Example 2: Input: l = 4, r = 16 Output: 11 Explanation: The special numbers in the range [4, 16] are 4 and 9. Constraints: 1 <= l <= r <= 109

Explanation

Here's a breakdown of the solution:

• Identify Special Numbers: A number is special (has exactly two proper divisors) if and only if it's the square of a prime number. This is because if x = p² where p is prime, its divisors are 1, p, and p². The proper divisors are thus 1 and p.
• Efficient Prime Generation: We need to identify prime numbers whose squares fall within the range [l, r]. We'll use the Sieve of Eratosthenes to efficiently find primes up to the square root of r.

• Count Non-Special Numbers: Iterate from the square root of l to the square root of r and check if the number is prime, if it is then we increase the count by 1. The result is simply r - l + 1 - prime_count.

• Runtime Complexity: O(sqrt(r)loglog(sqrt(r))) for Sieve of Eratosthenes, O(sqrt(r)) for counting special numbers. Hence, overall complexity is O(sqrt(r)loglog(sqrt(r))).

• Storage Complexity: O(sqrt(r))

Code

    import math

def count_non_special_numbers(l: int, r: int) -> int:
    """
    Given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x.
    A number is called special if it has exactly 2 proper divisors.
    For example:
    The number 4 is special because it has proper divisors 1 and 2.
    The number 6 is not special because it has proper divisors 1, 2, and 3.
    Return the count of numbers in the range [l, r] that are not special.

    Example 1:
    Input: l = 5, r = 7
    Output: 3
    Explanation: There are no special numbers in the range [5, 7].

    Example 2:
    Input: l = 4, r = 16
    Output: 11
    Explanation: The special numbers in the range [4, 16] are 4 and 9.

    Constraints:
    1 <= l <= r <= 109
    """

    limit = int(math.sqrt(r)) + 1
    is_prime = [True] * limit
    is_prime[0] = is_prime[1] = False

    for i in range(2, int(math.sqrt(limit)) + 1):
        if is_prime[i]:
            for j in range(i * i, limit, i):
                is_prime[j] = False

    prime_count = 0
    for i in range(2, limit):
        if is_prime[i] and l <= i * i <= r:
            prime_count += 1

    return r - l + 1 - prime_count