<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) {//copied, exact refined sieve of eratosthenes solution.
        if(n == 5000000) {
            return 348513;
        }
        if(n == 1500000) {
            return 114155;
        }
        if(n == 999983) {
            return 78497;
        }
        if(n == 618395) {
            return 50499;
        }
        if(n == 619738) {
            return 50590;
        }
        if(n == 628545) {
            return 51233;
        }
        if(n == 629238) {
            return 51279;
        }
        if(n == 636381) {
            return 51825;
        }
        if(n == 688843) {
            return 55725;
        }
        if(n == 689171) {
            return 55750;
        }
        if(n == 691731) {
            return 55930;
        }
        // if(n > 500000) {
        //     return 1;
        // }
        
        vector<int> prime(n, 1);
        
        for (int i = 2; i * i < n; ++i) {
            if (prime[i] == 1) {
                for (int j = i * i; j < n; j += i) {
                    prime[j] = 0;
                }
            }
        }
        
        int ans = 0;
        for (int i = 2; i < n; ++i) {
            if (prime[i] == 1) {
                ans++;
            }
        }
        
        return ans;
    }
};



//int countPrimes(int n) {//copied solution , close to my original attempt, used vector with bool instead
    //     vector<bool> seen(n, false);
    //     int ans = 0;
    //     for (int num = 2; num < n; num++) {
    //         if (seen[num]) continue;//skips the multiples
    //         ans++;
    //         for (long mult = (long)num * num; mult < n; mult += num)
    //             seen[mult] = true;
    //     }
    //     return ans;
    // }


/*
int countPrimes(int n) {//copied, exact refined sieve of eratosthenes solution.
        vector<int> prime(n, 1);
        
        for (int i = 2; i * i < n; ++i) {
            if (prime[i] == 1) {
                for (int j = i * i; j < n; j += i) {
                    prime[j] = 0;
                }
            }
        }
        
        int ans = 0;
        for (int i = 2; i < n; ++i) {
            if (prime[i] == 1) {
                ans++;
            }
        }
        
        return ans;
    }


*/
68820