Count Primes LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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In this post, you will find the solution for the Count Primes in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Count Primes– LeetCode Problem

Count Primes– LeetCode Problem

Problem:

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– LeetCode Solutions

Count Primes Solution in 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);
  }
};

Count Primes Solution in 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;
  }
}

Count Primes Solution in 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)

• Time: O(nloglogn)
• Space: O(n)