<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:
        def int_sqrtt(n):
            if n < 0:
                return -1
            if n == 0:
                return 0

            low, high = 1, n
            while low <= high:
                mid = (low + high) // 2
                sq=mid * mid
                if sq == n:
                    return mid
                elif sq < n:
                    low = mid + 1
                else:
                    high = mid - 1

            return high
        if n <= 2:
            return 0
        isPrime = np.ones(n>>1, dtype=np.byte)
        isPrime[0] = 0
        i = 1
        sq=int(math.sqrt(n))
        while (i<<1)+1 <= sq:
            if isPrime[i]:
                isPrime[2*i*i+2*i::2*i+1] = 0
            i += 1
        
        return np.sum(isPrime) + 1 # +1 for 2
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