<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 isPrime[10000010];

    void sieve(int n,vector<int>&prime){
        fill(isPrime,isPrime+n,1);
        isPrime[0]=isPrime[1]=0;
        for(long long i=2;i<=n;i++){
            if(isPrime[i]) prime.push_back(i);
            for(long long j=i*i;j<=n;j+=i){
                isPrime[j]=0;
            }
        }
    }
    int countPrimes(int n) {
        vector<int>prime;
        sieve(n,prime);
        // for(auto &it:prime) cout<<it<< " ";
        int count=lower_bound(begin(prime),end(prime),n)-begin(prime);

        return count;
    }
};
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