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Formatted question description: https://leetcode.ca/all/762.html

762. Prime Number of Set Bits in Binary Representation (Easy)

Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) 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 1s present when written in binary. For example, 21 written in binary is 10101 which has 3 set bits. Also, 1 is not a prime.)

Example 1:

Input: L = 6, R = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
9 -> 1001 (2 set bits , 2 is prime)
10->1010 (2 set bits , 2 is prime)

Example 2:

Input: L = 10, R = 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)

Note:

  1. L, R will be integers L <= R in the range [1, 10^6].
  2. R - L will be at most 10000.

Companies:
Amazon

Related Topics:
Bit Manipulation

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/
// Time: O(N)
// Space: O(1)
class Solution {
private:
    int getBitCount(int n) {
        int cnt = 0;
        while (n) {
            cnt += n % 2;
            n >>= 1;
        }
        return cnt;
    }
public:
    int countPrimeSetBits(int L, int R) {
        unordered_set<int> s{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 };
        int ans = 0;
        for (; L <= R; ++L) {
            int cnt = getBitCount(L);
            if (s.find(cnt) != s.end()) ++ans;
        }
        return ans;
    }
};

  • class Solution {
    public int countPrimeSetBits(int L, int R) {
        int count = 0;
        for (int i = L; i <= R; i++) {
            int ones = Integer.bitCount(i);
            if (isPrime(ones))
                count++;
        }
        return count;
    }

    public boolean isPrime(int num) {
        if (num == 1)
            return false;
        if (num == 2 || num == 3)
            return true;
        if (num % 2 == 0 || num % 3 == 0)
            return false;
        int upper = (int) Math.sqrt(num);
        for (int i = 5; i <= upper; i += 2) {
            if (num % i == 0)
                return false;
        }
        return true;
    }
}

  • // OJ: https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/
// Time: O(N)
// Space: O(1)
class Solution {
private:
    int getBitCount(int n) {
        int cnt = 0;
        while (n) {
            cnt += n % 2;
            n >>= 1;
        }
        return cnt;
    }
public:
    int countPrimeSetBits(int L, int R) {
        unordered_set<int> s{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 };
        int ans = 0;
        for (; L <= R; ++L) {
            int cnt = getBitCount(L);
            if (s.find(cnt) != s.end()) ++ans;
        }
        return ans;
    }
};

  • 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))

############

import math
class Solution(object):
    def isPrime(self, num):
        if num == 1:
            return 0
        elif num == 2:
            return 1
        for i in xrange(2, int(math.sqrt(num))+1):
            if num % i == 0:
                return 0
        return 1
    def countPrimeSetBits(self, L, R):
        """
        :type L: int
        :type R: int
        :rtype: int
        """
        return sum(self.isPrime(bin(num)[2:].count('1')) for num in xrange(L, R+1))

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