# 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.

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

• 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;
}
}

• 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;
}
};

• 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))


• 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
}