# Solving Leetcode Interviews in Seconds with AI: Count Primes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "204" 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 number of prime numbers that are strictly less than n.   Example 1:  Input: n = 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.  Example 2:  Input: n = 0 Output: 0  Example 3:  Input: n = 1 Output: 0    Constraints:  0 <= n <= 5 * 106  

	# Explanation
	*   **Sieve of Eratosthenes:** Use the Sieve of Eratosthenes algorithm to efficiently find all prime numbers less than `n`. This involves creating a boolean array representing numbers from 0 to `n-1`, initially marking all as potentially prime, and then iteratively marking the multiples of each prime number as composite (not prime).
*   **Optimization:** Start the marking of multiples from the square of the prime number. This significantly reduces redundant calculations because all smaller multiples would have already been marked by smaller prime numbers.
*   **Counting Primes:** After marking all composite numbers, simply count the number of remaining numbers that are marked as prime in the boolean array.

*   **Runtime Complexity:** O(n log log n), **Storage Complexity:** O(n)

	
	# Code
	```python
	def countPrimes(n):
    """
    Counts the number of prime numbers less than a non-negative number, n.

    Args:
        n: A non-negative integer.

    Returns:
        The number of prime numbers less than n.
    """
    if n <= 2:
        return 0

    isPrime = [True] * n
    isPrime[0] = isPrime[1] = False

    for i in range(2, int(n**0.5) + 1):
        if isPrime[i]:
            for j in range(i*i, n, i):
                isPrime[j] = False

    return sum(isPrime)
	```