<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 numbers[0] and numbers[1] as False because 0 and 1 are not prime.
        # Initialze numbers[2] through numbers[n-1] as True because we assume each number
        # is prime until we find a prime number (p) that is a divisor of the number
        numbers = [False, False] + [True] * (n - 2)
        for p in range(2, int(sqrt(n)) + 1):
            if numbers[p]:
                # Set all multiples of p to false because they are not prime.
                for multiple in range(p * p, n, p):
                    numbers[multiple] = False

        # numbers[index] will only be true where index is a prime number
        # return the number of indices whose value is true.
        return sum(numbers)
88056