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Prime Number of Set Bits in Binary Representation⚓︎

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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:
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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:
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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⚓︎

Way 1⚓︎

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class Solution {
private:
    bool isPrime(int x) {
        if (x < 2) return false;
        for (int i = 2; i <= x / i; i++) {
            if (x % i == 0) return false;
        }
        return true;
    }

    int bitCount(int x) {
        int cnt = 0;
        while (x) {
            cnt += x & 1;
            x >>= 1;
        }
        return cnt;
    }

public:
    int countPrimeSetBits(int left, int right) {
        int res = 0;
        for (int i = left; i <= right; i++) {
            if (isPrime(bitCount(i))) res++;
        }
        return res;
    }
};

Using the built-in function (see reference):

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class Solution {
private:
    bool isPrime(int x) {
        if (x < 2) return false;
        for (int i = 2; i <= x / i; i++) {
            if (x % i == 0) return false;
        }
        return true;
    }

public:
    int countPrimeSetBits(int left, int right) {
        int res = 0;
        for (int i = left; i <= right; i++) {
            if (isPrime(__builtin_popcount(i))) res++;
        }
        return res;
    }
};
  • Time complexity: \(O\left( \left( right-left \right) \cdot \sqrt{\log right} \right)\);
  • Space complexity: \(O(1)\).

Way 2⚓︎

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class Solution {
public:
    int countPrimeSetBits(int left, int right) {
        unordered_set<int> primes({2, 3, 5, 7, 11, 13, 17, 19});

        int res = 0;
        for (int i = left; i <= right; i++) {
            int s = 0;
            for (int k = i; k; k >>= 1) s += k & 1;
            if (primes.count(s)) res++;
        }
        return res;
    }
};

Way 3⚓︎

See reference.

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class Solution {
public:
    int countPrimeSetBits(int left, int right) {
        string bin("10100010100010101100");
        int mask = stoi(bin, nullptr, 2);

        int res = 0;
        for (int i = left; i <= right; i++) {
            if ((1 << __builtin_popcount(i)) & mask) res++;
        }

        return res;
    }
};
  • Time complexity: \(O(right-left)\);
  • Space complexity: \(O(1)\).