Massive Algorithms: LeetCode 204 - Count Primes (Java)

LeetCode – Count Primes (Java)
Count the number of prime numbers less than a non-negative number, n
This solution is the implementation of Sieve of Eratosthenes.

X. Using BitSet
https://leetcode.com/discuss/33528/accepted-java-code-with-bitset
java.util.BitSet.nextClearBit(int)

* Returns the index of the first bit that is set to {@code false}

* that occurs on or after the specified starting index.

java.util.BitSet.nextSetBit(int)

* Returns the index of the first bit that is set to {@code true}

* that occurs on or after the specified starting index. If no such

* bit exists then {@code -1} is returned.

java.util.BitSet.set(int, int)

* Sets the bits from the specified {@code fromIndex} (inclusive) to the

* specified {@code toIndex} (exclusive) to {@code true}.

java.util.BitSet.clear(int, int)

* Sets the bits from the specified {@code fromIndex} (inclusive) to the

* specified {@code toIndex} (exclusive) to {@code false}.

java.util.BitSet.cardinality()

* Returns the number of bits set to {@code true} in this {@code BitSet}.

public int countPrimes(int n) { BitSet bs = new BitSet(n); bs.set(0); bs.set(1); int ind = 0, count = 0; while(ind < n){ ind = bs.nextClearBit(ind + 1); if(ind >= n) return count; count++; for(int i = 2 * ind; i < n; i += ind) bs.set(i); } return count; }

public int countPrimes(int n) {
    if (n <= 1) {
        return 0;
    }

    BitSet set = new BitSet(n);
    set.set(0, 2);

    for (int i = 2; i < n; i++) {
        if (!set.get(i)) {
            for (int j = i*2; j < n; j+=i) {
                set.set(j);
            }    
        }

    }

    return n-set.cardinality();
}

https://gist.github.com/gcrfelix/4c11d5107f73ff1e33d5
// new method: 10/28
public int countPrimes(int n) {
if(n <= 2) return 0;
boolean[] notPrime = new boolean[n];
int count = 0;
for(int i=2; i<n; i++) {
if(notPrime[i] == false) {
count ++;
for(int j=2; i*j<n; j++) {
notPrime[i*j] = true;
}
}
}
return count;
}

public int countPrimes(int n) {
if(n <= 2) {
return 0;
}
int count = 1; // count = 1 because 2 is prime, add it to total first
for(int i=3; i<n; i++) {
if(isPrime(i)) {
count ++;
}
}
return count;
}

private boolean isPrime(int n) {
if(n%2 == 0) { // even numbers can't be prime except 2.
return false;
}
for(int i=3; i*i<=n; i+=2) { // We start from 3 and add 2 to avoid even numbers(we should get 3, 5, 7 etc.).
if(n%i == 0) {
return false;
}
}
return true;

}

http://algobox.org/count-primes/

The accurate time complexity is $O(n\log{\log{n}})$ which is not trivial to show. But, it is easy to show a complexity of $O(n\log{n})$ .

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

http://stackoverflow.com/questions/2582732/time-complexity-of-sieve-of-eratosthenes-algorithm

The complexity of the algorithm is O(n(logn)(loglogn)) bit operations.

How do you arrive at that?

That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere.

Suppose I am running the sieve on the first 100 numbers (n = 100), assuming that marking the numbers as composite takes constant time (array implementation), the number of times we use mark_composite() would be something like

n/2 + n/3 + n/5 + n/7 + ... + n/97        =      O(n^2)

And to find the next prime number (for example to jump to 7 after crossing out all the numbers that are multiples of 5), the number of operations would be O(n).

Your n/2 + n/3 + n/5 + … n/97 is not O(n^2), because the number of terms is not constant. [Edit after your edit: O(n²) is too loose an upper bound.] A loose upper-bound is n(1+1/2+1/3+1/4+1/5+1/6+…1/n) (sum of reciprocals of all numbers up to n), which is O(n log n): see Harmonic number. A more proper upper-bound is n(1/2 + 1/3 + 1/5 + 1/7 + …), that is sum of reciprocals of primes up to n, which is O(n log log n). (See here or here.)
The "find the next prime number" bit is only O(n) overall, amortized — you will move ahead to find the next number only n times in total, not per step. So this whole part of the algorithm takes only O(n).

So using these two you get an upper bound of O(n log log n) + O(n) = O(n log log n) arithmetic operations. If you count bit operations, since you're dealing with numbers up to n, they have about log n bits, which is where the factor of log n comes in, giving O(n log n log log n) bit operations.

public int countPrimes(int n) {
 if (n <= 2)
  return 0;
 
 // init an array to track prime numbers
 boolean[] primes = new boolean[n]; // we can use notPrimes
 for (int i = 2; i < n; i++)
  primes[i] = true;
 
 for (int i = 2; i <= Math.sqrt(n - 1); i++) {
 // or for (int i = 2; i <= n-1; i++) {
  if (primes[i]) {
   for (int j = i + i; j < n; j += i)
    primes[j] = false;
  }
 }
 
 int count = 0;
 for (int i = 2; i < n; i++) {
  if (primes[i])
   count++;
 }
 
 return count;
}

https://leetcode.com/discuss/35195/my-simple-java-solution
public int countPrimes(int n) { boolean[] notPrime = new boolean[n]; int count = 0; for (int i = 2; i < n; i++) { if (notPrime[i] == false) { count++; for (int j = 2; i*j < n; j++) { notPrime[i*j] = true; } } } return count; }
http://bookshadow.com/weblog/2015/04/27/leetcode-count-primes/
http://www.zhuangjingyang.com/leetcode-count-primes/
public int countPrimes(int n) { boolean notPrime[] = new boolean[n + 2]; notPrime[0] = notPrime[1] = true; for (int i = 2; i * i < n; i++) { if (!notPrime[i]) { int c = i * i; while (c < n) { notPrime[c] = true; c += i; } } } int ans = 0; for (int i = 0; i < n; i++) { if (!notPrime[i]) ans ++; } return ans; }

This solution exceeds time limit.

public int countPrimes(int n) {
    n = n-1;
 
    ArrayList<Integer> primes = new ArrayList<Integer>();
 
    if(n<=1) 
        return 0;
    if(n==2)
        return 1;
    if(n==3)
        return 2;
 
    primes.add(2);
    primes.add(3);
 
    for(int i=4; i<=n; i++){
        boolean isPrime = true;
        for(int p: primes){
            int m = i%p;
            if(m==0){
                isPrime = false;
                break;
            }
        }
 
        if(isPrime){
            primes.add(i);
        }
    }
 
    return primes.size();
}

http://blog.csdn.net/myself9711/article/details/45437229

Time Complexity: O(n*n)

class Solution:
    # @param {integer} n
    # @return {integer}
    def countPrimes(self, n):
        result_dict = {0: 0, 1: 1, 2: 2}
        for key in range(3, n+1):
            if self.isPrime(key):
                result_dict[key] = result_dict[key-1] + 1
            else:
                result_dict[key] = result_dict[key-1]
        return result_dict[n]


    def isPrime(self, n):
        if n == 0:
            return False
        if n == 1 or n == 2:
            return True
        for divide in range(2, n):
            if n % divide == 0:
                return False
        return True

http://www.programcreek.com/2014/04/leetcode-count-primes-java/

public int countPrimes(int n) {
    n = n-1;
 
    ArrayList<Integer> primes = new ArrayList<Integer>();
 
    if(n<=1) 
        return 0;
    if(n==2)
        return 1;
    if(n==3)
        return 2;
 
    primes.add(2);
    primes.add(3);
 
    for(int i=4; i<=n; i++){
        boolean isPrime = true;
        for(int p: primes){
            int m = i%p;
            if(m==0){
                isPrime = false;
                break;
            }
        }
 
        if(isPrime){
            primes.add(i);
        }
    }
 
    return primes.size();
}

Read full article from LeetCode – Count Primes (Java)

LeetCode 204 - Count Primes (Java)

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