Count of squarefree numbers in range

Question

Given two numbers x and y , find count of numbers that are squarefree where squarefree number is one divisible by no perfect square, except 1. For example, 10 is square-free but 18 is not, as it is divisible by 9 = 32. Few positive square-free numbers are :

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15 ...

limits

1 <= X,Y <= 10^9

0 <= |X-Y| <= 10^6

x=10 , Y=15 

gives

ans=5

My approach is to generate all prime till squareroot(10^9) (sieve of eratosthenes), and check whether each number in given range divisible by square of prime . Count of such numbers is substracted from length of range to give square free numbers .

But this approach time out in complexity , please suggest some other approach

Answer 1

Use the inclusion-exclusion principle:

Let f(n) = number of non-square-free numbers in 1 ... n. We will only do inclusion-exclusion on the squares of primes, so as to avoid overcounting for squares of squares.

We have:

f(n) = n / 4      => these are divisible by 4, so NOT square-free
       +
       n / 9      => these are divisible by 9, so NOT square-free
       + 
       n / 25     => these are divisible by 16, so NOT square-free
       +
       ...
       -
       n / (4*9)  => these are divisible by both 4 and 9, so we overcounted
       -
       n / (4*25) => these are divisible by both 4 and 25, so we overcounted 
       -
       ...

How efficient is this?

We only need primes p such that p^2 <= 10^9, meaning p < 31623. This already isn't a lot of primes, any trivial sieve (or maybe even trial division) should be fast enough. Then apply inclusion-exclusion, which will also be fast since the products of squared primes will get large fast so you'll be able to terminate prematurely in a a lot of cases (n / something = 0 whenever something > n).

In order to see why you'll be able to terminate prematurely, rewrite the above as:

f(n) = n / (2^2) -
       n / (2^2*3^2) +
       n / (2^2*3^2*5^2) -  <= this becomes 0 very fast. 
                               When it does, 
                               backtrack and increment the previous term.
                               For example, if this were 0,
                               you'd do - n / (2^2*5^2) next
       ...

More info on this here.

Answer 2

To expand on my comment: assume that X <= Y and initialize a boolean array SF[X..Y] to be all true. For every k from 2 to floor(sqrt(Y)) (optionally including the composites; the asymptotic running time stays the same), for every multiple m of k² between X and Y, set SF[m] to false. Return the number of true values remaining in SF.

The running time is O((Y - X) + sqrt(Y)), since Σ_k=1,...,∞ 1/k² = π²/6.