Description
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 <= 1060 <= right - left <= 104
Solutions
Solution 1: Math + Bit Manipulation
In the problem, both left and right are within the range of 10^6, and since 2^{20} = 1048576, the number of 1s in binary representation can be at most 20. The prime numbers within 20 are [2, 3, 5, 7, 11, 13, 17, 19].
We enumerate each number in the range [left,.. right], count the number of 1s in its binary representation, and then check if this count is a prime number. If it is, we increment the answer by one.
The time complexity is O(n× log m), where n = right - left + 1 and m is the maximum number in the range [left,.. right].
PythonJavaC++GoTypeScriptRustC#
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))(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
function countPrimeSetBits(left: number, right: number): number {
const primes = new Set<number>([2, 3, 5, 7, 11, 13, 17, 19]);
let ans = 0;
for (let i = left; i <= right; i++) {
const bits = bitCount(i);
if (primes.has(bits)) {
ans++;
}
}
return ans;
}
function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}(code-box)
impl Solution {
pub fn count_prime_set_bits(left: i32, right: i32) -> i32 {
let primes = [2, 3, 5, 7, 11, 13, 17, 19];
let mut ans = 0;
for i in left..=right {
let bits = i.count_ones() as i32;
if primes.contains(&bits) {
ans += 1;
}
}
ans
}
}(code-box)
public class Solution {
public int CountPrimeSetBits(int left, int right) {
var primes = new HashSet<int> { 2, 3, 5, 7, 11, 13, 17, 19 };
int ans = 0;
for (int i = left; i <= right; ++i) {
int bits = BitOperations.PopCount((uint)i);
if (primes.Contains(bits)) {
++ans;
}
}
return ans;
}
}(code-box)
