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 is10101
, which has3
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}
return sum(i.bit_count() in primes for i in range(left, right + 1))
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;
}
};
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
}