<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>

import numpy as np
class Solution:
    def countPrimes(self, n: int) -> int:
        if n <= 2:
            return 0
        # isPrime[i] = 1 if 2*i+1 is prime
        isPrime = np.ones(n // 2, dtype=np.byte)
        isPrime[0] = 0
        i = 1
        sqrtn = np.sqrt(n)
        while 2*i+1 <= sqrtn:
            if isPrime[i]:
                # we start from the square of 2i+1
                # which is 4i*i+4i+1
                # which is 2(2i*i+2i)+1
                # so we use 2i*i+2i as the index
                # write out what happens for n=10, n=20 to see why
                isPrime[2*i*(i+1)::2*i+1] = 0
            i += 1
        
        return np.sum(isPrime) + 1 # +1 for 2)
22393