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
class Solution:
def countPrimeSetBits(self, left: int, right: int) -> int:
primes={2,3,5,7,11,13,17,19}
ans=0
for i in range(left, right+1):
if bin(i)[2:].count("1") in primes:
ans+=1
return ans
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