# Codility 'CountSemiprimes' Solution

Martin Kysel · August 3, 2014

##### Short Problem Definition:

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

SemiPrimes

##### Complexity:

expected worst-case time complexity is `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 of 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):
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
``````