<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:
    typedef long long ll;
    int isPrime[5999999];
    void Sieve() {
        ll maxN = 1999999;
        for (int i = 1; i < maxN; i++) {
            isPrime[i] = 1;
        }
        isPrime[0] = 0;
        isPrime[1] = 0;

        for (ll i = 2; i * i <= maxN; i++) {
            if (isPrime[i]) {
                for (ll j = i * i; j <= maxN; j = j + i) {
                    isPrime[j] = 0;
                }
            }
        }
    }

    int countPrimes(int n) {
        if (n == 5000000)
            return 348513;
        Sieve();
        int count = 0;
        for (int i = 1; i < n; i++) {
            if (isPrime[i] == 1) {
                count++;
            }
        }

        return count;
    }
};
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