Count Primes LeetCode Solution | Easy Approach

Count Primes Given an integer n, return the number of prime numbers that are strictly less than n.

Example 1:

Input: n = 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

Example 2:

Input: n = 0
Output: 0

Example 3:

Input: n = 1
Output: 0

Constraints:

• 0 <= n <= 5 * 106

 Count Primes Solutions

✅Time: O(n log log n)
✅Space: O(n)

C++

class Solution {
 public:
  int countPrimes(int n) {
    if (n <= 2)
      return false;

    vector<bool> prime(n, true);
    prime[0] = false;
    prime[1] = false;

    for (int i = 0; i < sqrt(n); ++i)
      if (prime[i])
        for (int j = i * i; j < n; j += i)
          prime[j] = false;

    return count(begin(prime), end(prime), true);
  }
};

Java

class Solution {
  public int countPrimes(int n) {
    if (n <= 2)
      return 0;

    int ans = 0;
    boolean[] prime = new boolean[n];
    Arrays.fill(prime, 2, n, true);

    for (int i = 0; i < Math.sqrt(n); ++i)
      if (prime[i])
        for (int j = i * i; j < n; j += i)
          prime[j] = false;

    for (final boolean p : prime)
      if (p)
        ++ans;

    return ans;
  }
}

Python

class Solution:
  def countPrimes(self, n: int) -> int:
    if n <= 2:
      return 0

    isPrime = [False] * 2 + [True] * (n - 2)

    for i in range(2, int(n**0.5) + 1):
      if isPrime[i]:
        for j in range(i * i, n, i):
          isPrime[j] = False

    return sum(isPrime)
 

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