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762. Prime Number of Set Bits in Binary Representation

Description

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.

Β 

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.

Β 

Constraints:

  • 1 <= left <= right <= 106
  • 0 <= right - left <= 104

Solutions

Solution 1: Math + Bit Manipulation

In the problem, both \(\textit{left}\) and \(\textit{right}\) are within the range of \(10^6\), and since \(2^{20} = 1048576\), the number of \(1\)s in binary representation can be at most \(20\). The prime numbers within \(20\) are \([2, 3, 5, 7, 11, 13, 17, 19]\).

We enumerate each number in the range \([\textit{left},.. \textit{right}]\), count the number of \(1\)s in its binary representation, and then check if this count is a prime number. If it is, we increment the answer by one.

The time complexity is \(O(n\times \log m)\), where \(n = \textit{right} - \textit{left} + 1\) and \(m\) is the maximum number in the range \([\textit{left},.. \textit{right}]\).

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class Solution:
    def countPrimeSetBits(self, left: int, right: int) -> int:
        primes = {2, 3, 5, 7, 11, 13, 17, 19}
        return sum(i.bit_count() in primes for i in range(left, right + 1))

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class Solution {
    private static Set<Integer> primes = Set.of(2, 3, 5, 7, 11, 13, 17, 19);

    public int countPrimeSetBits(int left, int right) {
        int ans = 0;
        for (int i = left; i <= right; ++i) {
            if (primes.contains(Integer.bitCount(i))) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countPrimeSetBits(int left, int right) {
        unordered_set<int> primes{2, 3, 5, 7, 11, 13, 17, 19};
        int ans = 0;
        for (int i = left; i <= right; ++i) {
            ans += primes.count(__builtin_popcount(i));
        }
        return ans;
    }
};

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func countPrimeSetBits(left int, right int) (ans int) {
    primes := map[int]int{}
    for _, v := range []int{2, 3, 5, 7, 11, 13, 17, 19} {
        primes[v] = 1
    }
    for i := left; i <= right; i++ {
        ans += primes[bits.OnesCount(uint(i))]
    }
    return
}

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function countPrimeSetBits(left: number, right: number): number {
    const primes = new Set<number>([2, 3, 5, 7, 11, 13, 17, 19]);
    let ans = 0;

    for (let i = left; i <= right; i++) {
        const bits = bitCount(i);
        if (primes.has(bits)) {
            ans++;
        }
    }

    return ans;
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

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impl Solution {
    pub fn count_prime_set_bits(left: i32, right: i32) -> i32 {
        let primes = [2, 3, 5, 7, 11, 13, 17, 19];
        let mut ans = 0;

        for i in left..=right {
            let bits = i.count_ones() as i32;
            if primes.contains(&bits) {
                ans += 1;
            }
        }

        ans
    }
}

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public class Solution {
    public int CountPrimeSetBits(int left, int right) {
        var primes = new HashSet<int> { 2, 3, 5, 7, 11, 13, 17, 19 };
        int ans = 0;

        for (int i = left; i <= right; ++i) {
            int bits = BitOperations.PopCount((uint)i);
            if (primes.Contains(bits)) {
                ++ans;
            }
        }

        return ans;
    }
}

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