Sieve of Eratosthenes to solve a Leetcode problem

Problem: https://leetcode.com/problems/prime-number-of-set-bits-in-binary-representation/

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.

My first solution used Miller-Rabin to do the primality test. Even being super fast, doing that test for every single input leads to a timeout. Solution was to use the Sieve of Eratosthenes, cache the non-primes in a static variable, and use that cache during the checks. That solution worked but it was still very inefficient albeit passing the bar for leetcode. An improvement (which I have not implemented yet) would be to store the cumulative number of primes from 1..N, and then just use cumulative[b] - cumulative[a-1], caching all the solutions. Cheers, Marcelo

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Threading.Tasks;
using System.IO;
namespace LeetCode
{
 class Program
 {
 static void Main(string[] args)
 {
 Solution sol = new Solution();
 Console.WriteLine(sol.CountPrimeSetBits(Int32.Parse(args[0]), Int32.Parse(args[1])));
 }
 }
 public class Solution
 {
 private static Hashtable notPrimes = null;
 public int CountPrimeSetBits(int L, int R)
 {
 if (Solution.notPrimes == null)
 {
 Solution.notPrimes = new Hashtable();
 Solution.notPrimes.Add(1, true);
 int N = 1000000;
 for (int i = 2; i <= (int)(Math.Sqrt(N) + 1); i++)
 {
 for (int j = i; i * j <= N; j++)
 {
 if (!Solution.notPrimes.ContainsKey(i * j)) Solution.notPrimes.Add(i * j, true);
 }
 }
 }
 int total = 0;
 for (int i = L; i <= R; i++)
 {
 int temp = i;
 int count = 0;
 while (temp > 0)
 {
 count += (temp % 2);
 temp /= 2;
 }
 total = !Solution.notPrimes.ContainsKey(count) ? total + 1 : total;
 }
 return total;
 }
 private bool IsPrimeMillerRabin(BigInteger n)
 {
 //It does not work well for smaller numbers, hence this check
 int SMALL_NUMBER = 1000;
 if (n <= SMALL_NUMBER)
 {
 return IsPrime(n);
 }
 int MAX_WITNESS = 500;
 for (long i = 2; i <= MAX_WITNESS; i++)
 {
 if (IsPrime(i) && Witness(i, n) == 1)
 {
 return false;
 }
 }
 return true;
 }
 private BigInteger SqRtN(BigInteger N)
 {
 /*++
 * Using Newton Raphson method we calculate the
 * square root (N/g + g)/2
 */
 BigInteger rootN = N;
 int count = 0;
 int bitLength = 1;
 while (rootN / 2 != 0)
 {
 rootN /= 2;
 bitLength++;
 }
 bitLength = (bitLength + 1) / 2;
 rootN = N >> bitLength;
 BigInteger lastRoot = BigInteger.Zero;
 do
 {
 if (lastRoot > rootN)
 {
 if (count++ > 1000) // Work around for the bug where it gets into an infinite loop
 {
 return rootN;
 }
 }
 lastRoot = rootN;
 rootN = (BigInteger.Divide(N, rootN) + rootN) >> 1;
 }
 while (!((rootN ^ lastRoot).ToString() == "0"));
 return rootN;
 }
 private bool IsPrime(BigInteger n)
 {
 if (n <= 1)
 {
 return false;
 }
 if (n == 2)
 {
 return true;
 }
 if (n % 2 == 0)
 {
 return false;
 }
 for (int i = 3; i <= SqRtN(n) + 1; i += 2)
 {
 if (n % i == 0)
 {
 return false;
 }
 }
 return true;
 }
 private int Witness(long a, BigInteger n)
 {
 BigInteger t, u;
 BigInteger prev, curr = 0;
 BigInteger i;
 BigInteger lln = n;
 u = n / 2;
 t = 1;
 while (u % 2 == 0)
 {
 u /= 2;
 t++;
 }
 prev = BigInteger.ModPow(a, u, n);
 for (i = 1; i <= t; i++)
 {
 curr = BigInteger.ModPow(prev, 2, lln);
 if ((curr == 1) && (prev != 1) && (prev != lln - 1)) return 1;
 prev = curr;
 }
 if (curr != 1) return 1;
 return 0;
 }
 }
}

Comments

  1. Since the problem was marked as easy, I have decided to cheat and:
    1) use the fact that there are only a few prime numbers (2, 3, 5, 7, 11, 13, 17, 19) that we can encounter because it's the largest number of bits that can be set in a number from the range ending with 10^6.
    2) use a bit fiddling trick to count the number of set bits.

    The C++ solution is below:
    class Solution {
    private:
    inline int numberOfSetBits(int v) {
    v = v - ((v >> 1) & 0x55555555);
    v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
    return ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
    }
    public:
    int countPrimeSetBits(int L, int R) {
    bool is_prime[20] {false};
    is_prime[2] = is_prime[3] = is_prime[5] = is_prime[7] = is_prime[11] = is_prime[13] = is_prime[17] = is_prime[19] = true;
    int total = 0;
    for (int i = L; i <= R; ++i) total += is_prime[numberOfSetBits(i)];
    return total;
    }
    };

    And since I don't really understand the bit fiddling magic to make the bit counting O(1), I've decided to simplify things even further by using Rust:

    impl Solution {
    pub fn count_prime_set_bits(l: i32, r: i32) -> i32 {
    let mut is_prime: Vec = vec![false; 21];
    for prime in vec![2, 3, 5, 7, 11, 13, 17, 19] {
    is_prime[prime] = true;
    }
    let mut count = 0;
    for x in l..(r+1) {
    if is_prime[x.count_ones() as usize] {
    count += 1;
    }
    }
    count
    }
    }

    Cheers,
    Taras

    Reply Delete
    Replies

    1. in fact Rust solution can actually be made shorter:

      impl Solution {
      pub fn count_prime_set_bits(l: i32, r: i32) -> i32 {
      let mut is_prime: Vec = vec![false; 21];
      vec![2, 3, 5, 7, 11, 13, 17, 19].iter().for_each(|prime| is_prime[*prime as usize] = true);
      (l..(r+1)).filter(|x| is_prime[x.count_ones() as usize]).count() as i32
      }
      }

      Rust is awesome.

      Delete

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