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.
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 <= 1060 <= right - left <= 104class Solution {
public int countPrimeSetBits(int left, int right) {
int ans = 0;
for (int i = left; i <= right; i++) {
int bits = 0, x = i;
while (x > 0) {
bits += x & 1;
x = x >> 1;
}
if (isPrime(bits))
ans++;
}
return ans;
}
private boolean isPrime(int x) {
if (x == 1)
return false;
for (int i = 2; i * i <= x; i++)
if (x % i == 0)
return false;
return true;
}
}class Solution {
public int countPrimeSetBits(int left, int right) {
int ans = 0;
for (int i = left; i <= right; i++)
if (isPrime(Integer.bitCount(i)))
ans++;
return ans;
}
private boolean isPrime(int x) {
if (x == 1)
return false;
for (int i = 2; i * i <= x; i++)
if (x % i == 0)
return false;
return true;
}
}