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?
NOTE: 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.
I can see us doing a semi-automated approach here or we write a nice function. We probably write a nice function because we are geeks.
def single_digit_integer_to_spoken_language(n):
if n == 0:
return ""
assert(n > 0 and n < 10)
return {1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five',
6: 'six', 7: 'seven', 8: 'eight', 9: 'nine'}[n]
def double_digit_integer_to_spoken_language(n):
assert(n > 9 and n < 100)
try:
return {
10: 'ten', 11: 'eleven', 12: 'twelve', 13: 'thirteen',
14: 'fourteen', 15: 'fifteen', 16: 'sixteen',
17: 'seventeen', 18: 'eighteen', 19: 'nineteen'}[n]
except KeyError:
pass
a, b = str(n)
a = {2: 'twenty', 3: 'thirty', 4: 'forty', 5: 'fifty',
6: 'sixty', 7: 'seventy', 8: 'eighty', 9: 'ninety'}[int(a)]
b = integer_to_spoken_language(int(b))
return a + '-' + b
def triple_digit_integer_to_spoken_language(n):
a, b = str(n)[0], str(n)[1:]
a = single_digit_integer_to_spoken_language(int(a))
b = integer_to_spoken_language(int(b))
if not b:
return a + " hundred"
return a + " hundred and " + b
def four_digit_integer_to_spoken_language(n):
a, b = str(n)[0], str(n)[1:]
a = single_digit_integer_to_spoken_language(int(a))
b = integer_to_spoken_language(int(b))
return a + " thousand " + b
def integer_to_spoken_language(n):
l = len(str(n))
if l == 1:
return single_digit_integer_to_spoken_language(n)
elif l == 2:
return double_digit_integer_to_spoken_language(n)
elif l == 3:
return triple_digit_integer_to_spoken_language(n)
elif l == 4:
return four_digit_integer_to_spoken_language(n)
else:
raise Exception("Length not supported.")
assert(integer_to_spoken_language(5) == 'five')
assert(integer_to_spoken_language(19) == 'nineteen')
assert(integer_to_spoken_language(21) == 'twenty-one')
assert(integer_to_spoken_language(210) == 'two hundred and ten')
assert(integer_to_spoken_language(3000) == 'three thousand ')
assert(integer_to_spoken_language(8333) == 'eight thousand three hundred and thirty-three')
Okay, I won't win a code golf contest but at least we can get the solution now.
l = len("".join([integer_to_spoken_language(i) for i in range(1, 1001)]).replace(" ", "").replace("-", ""))
assert(l == 21124)
print(l)
Made the classical fourty/forty error I have already done four years ago. Some things don't change. Also I still do not really like this problem. The Haskell solution is actually way easier to read this time.