<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 {
private:
    void simpleSieve(int limit, vector<int>& primes) {
        vector<bool> isPrime(limit + 1, true);
        isPrime[0] = isPrime[1] = false;

        for (int i = 2; i <= limit; i++) {
            if (isPrime[i]) {
                primes.push_back(i);
                for (int j = i * i; j <= limit; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    int segmentedSieve(int L, int R) {
        if (L > R)
            return 0;
        int limit = sqrt(R);
        vector<int> primes;
        simpleSieve(limit, primes);

        vector<bool> isPrime(R - L + 1, true);

        for (int i = 0; i < primes.size(); i++) {
            int p = primes[i];

            // Calculate the first multiple of p in the range [L, R]
            int start = max(p * p, (L + p - 1) / p * p);

            for (int j = start; j <= R; j += p) {
                isPrime[j - L] = false;
            }
        }

        int cnt = 0;
        for (int i = L; i <= R; i++) {
            if (isPrime[i - L] && i != 1) {
                cnt++;
            }
        }
        return cnt;
    }

public:
    int countPrimes(int n) {
        if (n <= 2)
            return 0;
        return segmentedSieve(2, n - 1);
    }
};

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