<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>

from math import sqrt, ceil

class Solution:
    def countPrimes(self, n: int) -> int:
        if n <= 2:
            return 0
        # minus one cos 1 is not prime, plus one cos 2 is prime
        primeCount = (n // 2) - 1 + 1
        bitmask = [True] * (n // 2)
        bitmask[0] = False

        # skip 0 and 1 and 2 (since 2 is even)
        for i in range(3, ceil(sqrt(n)), 2):
            # check if not prime in mask
            isPrime = bitmask[(i - 1) // 2]
            if isPrime:
                for z in range(i*i, n, i):
                    if z % 2 == 1 and bitmask[(z - 1) // 2]:
                        primeCount -= 1
                        bitmask[(z - 1) // 2] = False
        return primeCount
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