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.)
Example 1:
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)
Example 2:
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)
Note:
-
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, R):
"""
:type L: int
:type R: int
:rtype: int
"""
def is_prime(a):
if a<=1:
return False
i=2
while i*i<=a:
if a%i==0:
return False
i+=1
return True ans=0
for i in range(L,R+1):
b=bin(i)[2:]
if is_prime(b.count('1')):
ans+=1 return ans