Python solution for google code jam 2013 Round 1C - Problem A - Consonants: https://code.google.com/codejam/contest/2437488/dashboard#s=p0

Consonants.py

"""
    Used to solve google code jam 2013 Round 1C - Problem A - Consonants
    Owner: Jamie Xu - [email protected]
"""
def Consonants():
    in_f = open('A-small-attempt0.in', 'r')
    out_f = open('output.txt', 'w')
    num_of_case = int(in_f.readline().rstrip('\n'))
    for i in range(1, num_of_case+1):
        solve_case(in_f, out_f, i)

N = 0
def solve_case(in_f, out_f, case_index):
    print "start handling case #{}".format(case_index)
    case_info = in_f.readline().rstrip('\n').split(" ")
    name = case_info[0]
    N = int(case_info[1])
    cleaned_name = clean_name(name)
    n = get_n(cleaned_name, N)
    out_f.write("Case #{}: {}\n".format(case_index, n))

def get_n(name, N):
    length = len(name)
    if length < N:
        return 0
    temp = 0
    for i in xrange(N, length+1):
        if is_valid(name[0:i], N):
            temp = temp + 1
    return temp + get_n(name[1:], N)

def is_valid(str, N):
    parts = str.split("1")
    for part in parts:
        if len(part) >= N:
            return True
    return False
    
def clean_name(name):
    return name.replace("a", "1").replace("e", "1").replace("i", "1").replace("o", "1").replace("u", "1")
    
if __name__ == '__main__':    
    Consonants()

Unfortunately, this only solved the small data set.

Current solution is calculating how many valid ones are starting with current character then add the total to the recursive function with current character being removed.

I was thinking about a potential faster way but failed to complete:

1. name_piece = name.split("1")
2. for piece in name_piece, calculate many valid ones are reaching the boarder of the piece, i.e. those having the ability to "grow" to the outside of the current piece. num_gained_from_growing will be something like len(name) - len(piece)
3. Remove duplicate ones.
   Failed to find a formular for this :<