Introduction

In this blog post, we will explore how to solve the LeetCode problem "762" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given two integers left and right, return the count of numbers in the inclusive range [left, right] having a prime number of set bits in their binary representation. Recall that the number of set bits an integer has is the number of 1's present when written in binary. For example, 21 written in binary is 10101, which has 3 set bits. Example 1: Input: left = 6, right = 10 Output: 4 Explanation: 6 -> 110 (2 set bits, 2 is prime) 7 -> 111 (3 set bits, 3 is prime) 8 -> 1000 (1 set bit, 1 is not prime) 9 -> 1001 (2 set bits, 2 is prime) 10 -> 1010 (2 set bits, 2 is prime) 4 numbers have a prime number of set bits. Example 2: Input: left = 10, right = 15 Output: 5 Explanation: 10 -> 1010 (2 set bits, 2 is prime) 11 -> 1011 (3 set bits, 3 is prime) 12 -> 1100 (2 set bits, 2 is prime) 13 -> 1101 (3 set bits, 3 is prime) 14 -> 1110 (3 set bits, 3 is prime) 15 -> 1111 (4 set bits, 4 is not prime) 5 numbers have a prime number of set bits. Constraints: 1 <= left <= right <= 106 0 <= right - left <= 104

Explanation

Here's an efficient solution to the problem:

• Precompute Primes: Generate a set of prime numbers up to the maximum possible number of set bits (20, since 10⁶ < 2²⁰ and the max number of set bits is therefore 20). Using a boolean array based sieve of Eratosthenes is an efficient way to do this.

• Iterate and Count Set Bits: Iterate through the numbers in the given range [left, right]. For each number, efficiently count its set bits using bit manipulation.

• Check Primality: Check if the set bit count is present in the precomputed prime set. If it is, increment the result counter.

• Runtime & Storage Complexity: O(N + (right - left)), O(1). Where N is the maximum number of bits (20). Because right-left can be up to 10^4, the for loop can take up to that time.

Code

    def count_prime_set_bits(left: int, right: int) -> int:
    """
    Counts the numbers in the inclusive range [left, right] having a prime number of set bits.
    """

    # Precompute primes up to 20 (maximum possible set bits).
    primes = {2, 3, 5, 7, 11, 13, 17, 19}

    count = 0
    for num in range(left, right + 1):
        set_bits = bin(num).count('1')  # Efficiently count set bits
        if set_bits in primes:
            count += 1

    return count