##### Short Problem Definition:

Count the semiprime numbers in the given range [a..b]

##### Link

##### Complexity:

expected worst-case time complexity isÂ O(N*log(log(N))+M);

expected worst-case space complexity is O(N+M)

##### Execution:

First get all semiprimes from an adaptation of the Sieve ofÂ Eratosthenes. Because we will be computing the difference many times a prefix sum is adequate. Get the number of semiprimes up to the point. The index P is decreased by 1 because we want to know all primes that start from P.

##### Solution:

def sieve(N): semi = set() sieve = [True]* (N+1) sieve[0] = sieve[1] = False i = 2 while (i*i <= N): if sieve[i] == True: for j in xrange(i*i, N+1, i): sieve[j] = False i += 1 i = 2 while (i*i <= N): if sieve[i] == True: for j in xrange(i*i, N+1, i): if (j % i == 0 and sieve[j/i] == True): semi.add(j) i += 1 return semi def solution(N, P, Q): semi_set = sieve(N) prefix = [] prefix.append(0) # 0 prefix.append(0) # 1 prefix.append(0) # 2 prefix.append(0) # 3 prefix.append(1) # 4 for idx in xrange(5, max(Q)+1): if idx in semi_set: prefix.append(prefix[-1]+1) else: prefix.append(prefix[-1]) solution = [] for idx in xrange(len(Q)): solution.append(prefix[Q[idx]] - prefix[P[idx]-1]) return solution

If you enjoyed this post, then make sure you subscribe to my Newsletter and/or Feed.