Problem

Given two integers left and right, return _thecount of numbers in the inclusive range _[left, right]having aprime 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.

Examples

Example 1

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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

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10

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 <= 10^6
  • 0 <= right - left <= 10^4

Solution

Intuition

For each number in the range [left, right], count the number of set bits (1s) in its binary representation. If this count is a prime number, increment the answer. Since the maximum number of set bits for numbers up to 10^6 is 20, we can precompute all primes up to 20.

Approach

  1. Precompute the set of primes up to 20.
  2. For each number in [left, right], count its set bits (using built-in or bitwise method).
  3. If the count is in the set of primes, increment the answer.
  4. Return the answer.

Code

C++
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#include <vector>
#include <unordered_set>
using namespace std;
int countPrimeSetBits(int left, int right) {
    unordered_set<int> primes = {2,3,5,7,11,13,17,19};
    int ans = 0;
    for (int x = left; x <= right; ++x) {
        int cnt = __builtin_popcount(x);
        if (primes.count(cnt)) ++ans;
    }
    return ans;
}
Go
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func countPrimeSetBits(left int, right int) int {
    primes := map[int]bool{2:true,3:true,5:true,7:true,11:true,13:true,17:true,19:true}
    ans := 0
    for x := left; x <= right; x++ {
        cnt := 0
        y := x
        for y > 0 {
            cnt += y & 1
            y >>= 1
        }
        if primes[cnt] {
            ans++
        }
    }
    return ans
}
Java
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import java.util.*;
class Solution {
    public int countPrimeSetBits(int left, int right) {
        Set<Integer> primes = Set.of(2,3,5,7,11,13,17,19);
        int ans = 0;
        for (int x = left; x <= right; ++x) {
            int cnt = Integer.bitCount(x);
            if (primes.contains(cnt)) ++ans;
        }
        return ans;
    }
}
Kotlin
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fun countPrimeSetBits(left: Int, right: Int): Int {
    val primes = setOf(2,3,5,7,11,13,17,19)
    var ans = 0
    for (x in left..right) {
        val cnt = Integer.bitCount(x)
        if (cnt in primes) ans++
    }
    return ans
}
Python
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def countPrimeSetBits(left: int, right: int) -> int:
    primes = {2,3,5,7,11,13,17,19}
    ans = 0
    for x in range(left, right+1):
        cnt = bin(x).count('1')
        if cnt in primes:
            ans += 1
    return ans
Rust
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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 x in left..=right {
        let cnt = x.count_ones();
        if primes.contains(&(cnt as i32)) {
            ans += 1;
        }
    }
    ans
}
TypeScript
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export function countPrimeSetBits(left: number, right: number): number {
    const primes = new Set([2,3,5,7,11,13,17,19]);
    let ans = 0;
    for (let x = left; x <= right; ++x) {
        let cnt = 0, y = x;
        while (y > 0) {
            cnt += y & 1;
            y >>= 1;
        }
        if (primes.has(cnt)) ans++;
    }
    return ans;
}

Complexity

  • ⏰ Time complexity: O(N log M), where N = right - left + 1, M is the maximum value (up to 10^6), log M for counting set bits.
  • 🧺 Space complexity: O(1) (ignoring output and the set of small primes).