Problem 17

Solution to Problem 17. Solved in Python.

Problem: If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.

Note: This is obviously not an optimal solution. Pure brute force with some repeated code…I was feeling a bit lazy. I’ll come back to this problem some day and write a better solution.

def special_cases(number):
    hashmap = {
        10: 'ten',
        11: 'eleven',
        12: 'twelve',
        13: 'thirteen',
        14: 'fourteen',
        15: 'fifteen',
        16: 'sixteen',
        17: 'seventeen',
        18: 'eighteen',
        19: 'nineteen'
    }

    if number in hashmap.keys():
        return hashmap[number]
    return None


def ones(number):
    hashmap = {
        0: 'zero',
        1: 'one',
        2: 'two',
        3: 'three',
        4: 'four',
        5: 'five',
        6: 'six',
        7: 'seven',
        8: 'eight',
        9: 'nine'
    }

    if number in hashmap.keys():
        return hashmap[number]
    return None


def tens(number):
    if 20 <= number < 30:
        return 'twenty'
    if 30 <= number < 40:
        return 'thirty'
    if 40 <= number < 50:
        return 'forty'
    if 50 <= number < 60:
        return 'fifty'
    if 60 <= number < 70:
        return 'sixty'
    if 70 <= number < 80:
        return 'seventy'
    if 80 <= number < 90:
        return 'eighty'
    if 90 <= number < 100:
        return 'ninety'
    return None


def more(number):
    if 100 <= number < 1000:
        return 'hundred'
    if 1000 <= number < 1000000:
        return 'million'
    return None

string = ''
for i in range(1, 1000):
    length_of_number = len(str(i))
    if(length_of_number == 1):
        string += ones(i)
    elif(length_of_number == 2):
        test_tens = special_cases(i)
        if(test_tens != None):
            string += test_tens
        else:
            if(i % 10 == 0):
                string += tens(i)
            else:
                string += tens(i) + str(ones(i % 10))
    elif(length_of_number == 3):
        first_digit = int(str(i)[0])
        string += ones(first_digit) + str(more(i))

        base = first_digit * 100
        tens_num = i - base

        if(tens_num >= 1 and tens_num < 10):
            string += 'and' + str(ones(tens_num))
        elif(tens_num >= 10 and tens_num < 20):
            string += 'and' + str(special_cases(tens_num))
        elif(tens_num >= 20):
            string += 'and'
            if(tens_num % 10 == 0):
                string += tens(tens_num)
            else:
                string += tens(tens_num) + str(ones(tens_num % 10))

print len(string + 'onethousand')