LeetCode-in-Java

762. Prime Number of Set Bits in Binary Representation

Easy

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 <= 106
• 0 <= right - left <= 104

Solution

public class Solution {
    public int countPrimeSetBits(int left, int right) {
        int count = 0;
        boolean[] notPrime = new boolean[33];
        notPrime[0] = true;
        notPrime[1] = true;
        sieve(notPrime);
        for (int i = left; i <= right; i++) {
            int num = i;
            int setBits = 0;
            while (num > 0) {
                num = num & (num - 1);
                setBits++;
            }
            if (!notPrime[setBits]) {
                count++;
            }
        }
        return count;
    }

    private void sieve(boolean[] notPrime) {
        for (int i = 2; i <= 32; i++) {
            if (!notPrime[i]) {
                for (int j = 2 * i; j <= 32; j += i) {
                    notPrime[j] = true;
                }
            }
        }
    }
}

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