<p>Given an integer <code>n</code>, return <em>the number of prime numbers that are strictly less than</em> <code>n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 5 * 10<sup>6</sup></code></li>
</ul>

class Solution:
    def countPrimes(self, n: int) -> int:
        if n <= 2:
            return 0

        # Initialize the is_prime list, considering only odd indices.
        is_prime = [True] * ((n // 2))
        
        # Mark 2 as prime.
        primes = [2]
        
        # Iterate over odd numbers starting from 3.
        for i in range(3, n, 2):
            if is_prime[i // 2]:  # i is prime if is_prime[i // 2] is True.
                primes.append(i)
                # Mark all multiples of i as non-prime.
                for j in range(i * i, n, 2 * i):
                    is_prime[j // 2] = False

        return len(primes)

# Example usage:
# n = 30
# print(sieve_of_eratosthenes_optimized(n))




        
37843