Brute Force Solution

The most straightforward approach is to iterate through each number in the given range [left, right], calculate the number of set bits (1s) in its binary representation, and then check if that count is a prime number. If it is, we increment our counter.

Algorithm

1. Iterate through the numbers from left to right (inclusive).
2. For each number, calculate the number of set bits.
3. Check if the number of set bits is prime.
4. If it is, increment the count.
5. Return the final count.

Code (Python)

def count_prime_set_bits_brute_force(left: int, right: int) -> int:
    def count_set_bits(n: int) -> int:
        count = 0
        while n > 0:
            n &= (n - 1)
            count += 1
        return count

    def is_prime(n: int) -> bool:
        if n <= 1:
            return False
        for i in range(2, int(n**0.5) + 1):
            if n % i == 0:
                return False
        return True

    count = 0
    for i in range(left, right + 1):
        set_bits = count_set_bits(i)
        if is_prime(set_bits):
            count += 1
    return count

Time and Space Complexity

• Time Complexity: O((right - left) * (log(right) + sqrt(log(right)))), where right - left is the range of numbers, log(right) is the time to count set bits, and sqrt(log(right)) is the time to check for primality (since the maximum number of set bits will be log(right)).
• Space Complexity: O(1). Constant extra space is used.

Optimized Solution

We can optimize the primality check by pre-calculating a set of prime numbers up to the maximum possible number of set bits (which is the number of bits in the binary representation of right). Since right <= 10^6, the maximum number of set bits is at most 20. We can create a set of primes and then simply check if the number of set bits is present in our prime set. Also instead of computing square root for each is_prime call we can precompute the primes upto 20 since max number of bits is 20.

Algorithm

1. Create a set of prime numbers less than or equal to 20.
2. Iterate through the numbers from left to right (inclusive).
3. For each number, calculate the number of set bits.
4. Check if the number of set bits is in the pre-calculated prime number set.
5. If it is, increment the count.
6. Return the final count.

Code (Python)

def count_prime_set_bits_optimized(left: int, right: int) -> int:
    def count_set_bits(n: int) -> int:
        count = 0
        while n > 0:
            n &= (n - 1)
            count += 1
        return count

    primes = {2, 3, 5, 7, 11, 13, 17, 19}

    count = 0
    for i in range(left, right + 1):
        set_bits = count_set_bits(i)
        if set_bits in primes:
            count += 1
    return count

Time and Space Complexity

• Time Complexity: O((right - left) * log(right)). Iterating through the range takes O(right - left), and counting set bits takes O(log(right)). The primality check is now O(1).
• Space Complexity: O(1). We are only using a fixed size set of prime numbers.

Edge Cases

• If left or right are negative, the bit counting logic might have to be changed depending on the language.
• If left > right, then the for loop will not run, so return 0.