Integer to English Words

number_to_words.py

"""
Description:
Convert a non-negative integer to its english words representation. Given input is
guaranteed to be less than 2**31 - 1.

Examples:
Input: 123
Output: "One Hundred Twenty Three"

Input: 12345
Output: "Twelve Thousand Three Hundred Forty Five"

Input: 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand
         Eight Hundred Ninety One"
"""

def numberToWords(num):
    """
    :type num: int
    :rtype: str
    """
    base_dict = {
    0: ('', 'Ten'),
    1: ('One', 'Eleven'),
    2: ('Two', 'Twelve', 'Twenty'),
    3: ('Three', 'Thirteen', 'Thirty'),
    4: ('Four', 'Fourteen', 'Forty'),
    5: ('Five', 'Fifteen', 'Fifty'),
    6: ('Six', 'Sixteen', 'Sixty'),
    7: ('Seven', 'Seventeen', 'Seventy'),
    8: ('Eight', 'Eighteen', 'Eighty'),
    9: ('Nine', 'Nineteen', 'Ninety')
    }

    orders = [
        '', ' Thousand', ' Million', ' Billion', ' Trillion'
    ]

    def conv_2digits(v):
        tens, ones = divmod(v, 10)
        if tens == 0:
            return base_dict[ones][0]
        elif tens == 1:
            return base_dict[ones][1]
        else:
            if base_dict[ones][0]:
                return base_dict[tens][2] + ' ' + base_dict[ones][0]
            else:
                return base_dict[tens][2]

    def conv_3digits(v):
        # 3 digits
        hundreds, tens = divmod(v, 100)
        if hundreds == 0:
            return conv_2digits(tens)
        word = base_dict[hundreds][0] + ' Hundred'
        if tens == 0:
            return word
        else:
            return word + ' ' + conv_2digits(tens)

    if num == 0:
        return 'Zero'
    cur = num
    word = ''
    order = 0
    while True:
        if cur == 0:
            break
        cur, m = divmod(cur, 1000)
        v = conv_3digits(m)
        if v:
            v += orders[order]
            if word:
                word = v + ' ' + word
            else:
                word = v
        order += 1
    return word