<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>


vector<int> sieve(int n)
{
    vector<int> primeNumbers;
    if (n == 1 || n <= 0) return primeNumbers;

    const int idx = 1e8;
    bitset<idx> isprime;
    for (int i = 3; i <= n; i += 2)
        isprime[i] = 1;
    int sq = sqrt(n);
    for (int i = 3; i <= sq; i += 2)
        if (isprime[i])
            for (int j = i * i; j <= n; j += i)
                isprime[j] = 0;
    primeNumbers.push_back(2);
    for (int i = 3; i <= n; i += 2)
        if (isprime[i])
            primeNumbers.push_back(i);
    return primeNumbers;
}

class Solution
{
public:
    int countPrimes(int n)
    {
        vector<int> primes = sieve(n-1);
        return primes.size();
    }
};
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