Solution Explanation: Count Primes

This problem asks to find the number of prime numbers less than a given integer n. The most efficient approach is using the Sieve of Eratosthenes algorithm.

Algorithm (Sieve of Eratosthenes):

1. Initialization: Create a boolean array primes of size n, initially filled with True. primes[i] will be True if i is prime, and False otherwise. We don't need to consider 0 and 1 as they are not primes.

2. Iteration: Iterate from i = 2 up to sqrt(n). We only need to iterate up to the square root because if a number x has a divisor greater than sqrt(x), it must also have a divisor smaller than sqrt(x).

3. Marking Non-Primes: If primes[i] is True (meaning i is prime), mark all multiples of i as False (they are not prime). Start marking from i * i because smaller multiples would have already been marked by previous iterations.

4. Counting Primes: After the iteration, count the number of True values in the primes array (excluding indices 0 and 1). This count represents the number of primes less than n.

Time Complexity: O(n log log n). The outer loop iterates up to sqrt(n). The inner loop iterates through multiples of i, and the total number of iterations in the inner loop across all outer iterations is approximately proportional to n log log n which is significantly less than O(n).

Space Complexity: O(n). This is due to the boolean array primes of size n.

Code Explanation (Python):

class Solution:
    def countPrimes(self, n: int) -> int:
        primes = [True] * n  # Initialize boolean array
        ans = 0  # Initialize prime count
 
        for i in range(2, int(math.sqrt(n)) + 1): # Iterate up to sqrt(n)
            if primes[i]:  # If i is prime
                ans += 1  # Increment prime count
                for j in range(i * i, n, i):  # Mark multiples of i as non-prime
                    primes[j] = False
 
        #Count primes from 2 to sqrt(n)  and add primes between sqrt(n) and n if any.
        for i in range(int(math.sqrt(n)) + 1,n):
            if primes[i]:
                ans+=1
        return ans
 

The code in other languages (Java, C++, Go, JavaScript, C#) follows the same algorithm and logic. They differ only in syntax and specific library functions used. The core idea remains the efficient Sieve of Eratosthenes.

Example:

Let's say n = 10.

1. primes is initialized as [True, True, True, True, True, True, True, True, True, True]
2. The outer loop iterates from i = 2 to 3 (sqrt(10) ≈ 3.16).
3. When i = 2, multiples of 2 (4, 6, 8) are marked as False.
4. When i = 3, multiples of 3 (9) are marked as False.
5. Finally, primes becomes [False, False, True, True, False, True, False, True, False, False].
6. The number of True values (excluding indices 0 and 1) is 4 (2, 3, 5, 7), which is the correct answer.