7_Facebook_Decoding_porblem.py

'''
Given the mapping a = 1, b = 2, ... z = 26, and an encoded message, count the number of ways it can be decoded.

For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.

You can assume that the messages are decodable. For example, '001' is not allowed.
'''

char_map = {}
char_map['1'] = 'a'
char_map['2'] = 'b'
char_map['3'] = 'c'
char_map['4'] = 'd'
char_map['5'] = 'e'
char_map['6'] = 'f'
char_map['7'] = 'g'
char_map['8'] = 'h'
char_map['9'] = 'i'
char_map['10'] = 'j'
char_map['11'] = 'k'
char_map['12'] = 'l'
char_map['13'] = 'm'
char_map['14'] = 'n'
char_map['15'] = 'o'
char_map['16'] = 'p'
char_map['17'] = 'q'
char_map['18'] = 'r'
char_map['19'] = 's'
char_map['20'] = 't'
char_map['21'] = 'u'
char_map['22'] = 'v'
char_map['23'] = 'w'
char_map['24'] = 'x'
char_map['25'] = 'y'
char_map['26'] = 'z'


memo_dict = {} # stores the 'code => key' and 'number of ways => value' eg. '11'=> 2 ways(aa,k)
def count_ways(s):
    str_len = s.__len__()
    if s in memo_dict:
        return memo_dict[s]

    if str_len == 0:
        return 1

    if s[0] == '0':
        return 0


    if str_len > 0 and s[0] in char_map:
        memo_dict[s] = count_ways(s[1:])
    if str_len > 1 and s[:2] in char_map:
        memo_dict[s] += count_ways(s[2:])

    return memo_dict[s]


def coding_problem_7(s):
    """
    Given the mapping a = 1, b = 2, ... z = 26, and an encoded message, count the number of ways it can be decoded.
    Examples:
    >>> coding_problem_7('111')  # possible interpretations: 'aaa', 'ka', 'ak'
    3
    >>> coding_problem_7('2626')  # 'zz', 'zbf', 'bfz', 'bfbf'
    4
    """
    symbols = map(str, range(1, 27))
    if not s:
        return 1

    matches = filter(lambda symbol: s.startswith(symbol), symbols)
    encodings = [coding_problem_7(s[len(m):]) for m in matches]
    return sum(encodings)


if __name__ == '__main__':
    s =input()
    print("\n")
    print(count_ways(s))
    print(coding_problem_7(s))