Given two integers L
and R
, find the count of numbers in the range [L, R]
(inclusive) 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. Also, 1 is not a prime.)
Input: L = 6, R = 10 Output: 4 Explanation: 6 -> 110 (2 set bits, 2 is prime) 7 -> 111 (3 set bits, 3 is prime) 9 -> 1001 (2 set bits , 2 is prime) 10->1010 (2 set bits , 2 is prime)
Input: L = 10, R = 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)
L, R
will be integersL <= R
in the range[1, 10^6]
.R - L
will be at most 10000.
class Solution:
def countPrimeSetBits(self, L: int, R: int) -> int:
cnt = 0
for i in range(L, R + 1):
if bin(i).count('1') in {2, 3, 5, 7, 11, 13, 17, 19}:
cnt += 1
return cnt