<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 {
public:
    int countPrimes(int n) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        if(n == 0) return 0;
        n--;
        int s = 10000;
        vector<int> primes;
        // cout << n << endl;
        int nsqrt = sqrt(n);
        vector<char> is_primes(nsqrt+2, true);
        for(int i=2; i<=nsqrt; i++){
            if(is_primes[i]){
                primes.push_back(i);
                for(int j=i*i; j<=nsqrt; j+=i){
                    is_primes[j] = false;
                }
            }
        }

        int result = 0;
        vector<int> block(s);
        for(int k=0; k*s <= n; k++){
            fill(block.begin(), block.end(), true);
            int start = k*s;
            for(int p: primes){
                int start_indx = (start+p-1)/p;
                int j = max(start_indx, p)*p-start;
                for(; j<s ; j+=p){
                    block[j] = false;
                }
            }
            if(k==0) block[0] = block[1] = false;
            for (int i = 0;( (i < s) && (start + i <= n)); i++) {
                // cout << block[i] << " " << start+i << endl;
                if (block[i])
                    result++;
            }
        }

        return result;
    }
};
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