<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:
        # Seive of Eratosthnese
        # prime = [True for i in range(n)]
        # p = 2 
        # while p*p <= n:
        #     if prime[p]:
        #         for i in range(p*p, n, p):
        #             prime[i] = False
        #     p += 1
        # print(prime)
        # return sum([1 for i in prime if i==False])
        prime = [True for i in range(n+1)]
        p = 2
        while (p * p <= n):
    
            # If prime[p] is not
            # changed, then it is a prime
            if (prime[p] == True):
    
                # Update all multiples of p
                for i in range(p * p, n+1, p):
                    prime[i] = False
            p += 1
        count = 0 
        # Print all prime numbers
        for p in range(2, n):
            if prime[p]:
                print(p)
                count +=1
        return count 
70031