Count numbers with exactly 3 divisors

Given a number n, print count of numbers in the range from 1 to n having exactly3 divisors. 

Examples:

Input: n = 16
Output: 2
Explanation: Only 4 and 9 have exactly three divisors.

Input: n = 100
Output: 4
Explanation: 4, 9, 25 and 49 have exactly three divisors.

[Naive Approach] Nested Loop Method O(n^2) and Space O(1)

For each number, it counts how many divisors it has by checking all possible divisors from 1 to that number. If a number has exactly 3 divisors, it is printed

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[Expected Approach-1] Mathematical Approach

The number having exactly 3 divisors must be a perfect square of a prime number.

• A number with divisors is a number that can be evenly divided by 1, itself, and other integers in between.
• For a number to have exactly 3 divisors, the divisors must be:
  1. 1 (every number is divisible by 1)
  2. The number itself (a number is always divisible by itself)
  3. One other divisor between 1 and the number.

For this to happen, the number must be the square of a prime.

If p is a prime number, then the divisors of p² (the square of p) will be:
1 (as it's divisible by 1)
p (as it is divisible by p)
p² (as it is divisible by itself)

Therefore, the only way for a number to have exactly 3 divisors is if it is the square of a prime number. This is because, for any non-square number, there are more than 3 divisors, and for a perfect square of a non-prime number, it would have more than 3 divisors as well. Thus, the number must be the square of a prime, and that's why we specifically look at perfect squares of primes.

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Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)

[Expected Approach-2] Using Constant Space

In this method, we iterate through all numbers up to sqrt(n) and check if each is prime. If a number is prime, its square will have exactly three divisors (1, the prime itself, and its square). We count such primes whose squares are ≤ n. This approach avoids using extra space like a sieve, but has a time complexity of approximately O(n^(3/4)), because we perform a primality check (which takes up to O(√i) time) for each number up to √n.

Run a loop from 2 to sqrt(n) and check if the current element is prime or not, if it is so then print that number, but this method will increase the time complexity of the solution

Step for implementation:

• Start a loop for integer i from 2 to n.
• Check if i is prime or not, which can be done easily using the isPrime(n) method.
• If i is prime, check if its square is less than or equal to the given number. This will be reviewed only for squares of prime numbers, therefore reducing the number of checks.
• If the above condition is satisfied, the number will be printed and the loop will continue till i <= n.

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Time Complexity: O(n^3/4) because we call isPrime √n times, and for each i, isPrime(i) takes O(√i) time. Therefore, total time is the sum √1 + √2 + ... + √√n, which integrates to O(n^3/4)
Auxiliary Space: O(1)