Count Primes

Problem

Description:

Count the number of prime numbers less than a non-negative number, n.

Solution

Marking NotPrime Numbers: O(n lgn) time

public class Solution {
    public int countPrimes(int n) {
        if (n < 3) return 0;
        boolean[] notPrime = new boolean[n];
        int count = 0;
        for (int i = 2; i < n; i++) {
            if (!notPrime[i]) {
                count++;
                if (i > Math.sqrt(n)) continue;
                for (int j = 2; j * i < n; j++) notPrime[j*i] = true;
            }
        }
        return count;
    }
}

Analysis

The key idea in this solution is that: If a number is prime, then it's multiples are not primes
Therefore, we check from the first number 2
If we find a prime number, we increment count++
We avoid overflow by continuing the loop if i > Math.sqrt(n)
Then we mark multiples of current number as not prime in another for loop
We have notPrime so that we don't have to fill array after init