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762. Prime Number of Set Bits in Binary Representation

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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.

 

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

Solutions

Python3

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))

Java

class Solution {
    private static Set<Integer> primes = new HashSet<>(Arrays.asList(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;
    }
}

C++

class Solution {
public:
    unordered_set<int> primes{2, 3, 5, 7, 11, 13, 17, 19};

    int countPrimeSetBits(int left, int right) {
        int ans = 0;
        for (int i = left; i <= right; ++i)
            if (primes.count(__builtin_popcount(i)))
                ++ans;
        return ans;
    }
};

Go

func countPrimeSetBits(left int, right int) int {
	primes := map[int]bool{2: true, 3: true, 5: true, 7: true, 11: true, 13: true, 17: true, 19: true}
	ans := 0
	for i := left; i <= right; i++ {
		if primes[bits.OnesCount(uint(i))] {
			ans++
		}
	}
	return ans
}

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