13
\$\begingroup\$

Inspired by Project Euler #17, this is your challenge. Write a full program or function that takes a number as input, then print or return how many letters it would take to count up to and including that number in English (starting at one). You do not include spaces, commas, or hyphens, but you should include the word and. For example. 342 is spelled: Three Hundred and Forty-Two. This is 23 letters long.

Your input will be a positive integer. You do not have to handle invalid inputs. Built-ins or libraries that convert numbers to English are not allowed.

Here are all of the rules for how to spell numbers. (Note: I realize that some people use a different set of rules for how to spell numbers. This will just be the official rules for the purpose of this challenge)

1 to 20

one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen, twenty

21 to 99

Join these:

Twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety

to these:

-one, -two, -three, -four, -five, -six, -seven, -eight, -nine,

Note that four has a u but forty does not!

Examples:

53: Fifty-three 60: sixty 72: seventy-two 99: ninety-nine

100 to 999

Write out how many hundreds (one hundred, two hundred, three hundred, etc.), an "and", and the rest of the number as above. The and does count towards your letter score.

Examples:

101: One hundred and one 116: One hundred and sixteen 144: One hundred and forty-four 212: Two hundred and twelve 621: Six Hundred and twenty-one

1,000 to 999,999

Write how many thousands (one thousand, two thousand, etc.), a comma, then the rest of the number as above. Note that if you have no hundreds, you still need the and.

Examples:

1,101: One thousand, one hundred and one 15,016: Fifteen thousand and sixteen 362,928: Three hundred and sixty-two thousand, nine hundred and twenty-eight

Millions

Write out how many millions, then the rest of the number as above. Note that "A million" is 6 zeroes "1,000,000".

Examples:

191,232,891: One hundred and ninety-one million, two hundred and thirty-two thousand, eight hundred and ninety-one 1,006,101: One million, six thousand, one hundred and one

The same rule applies to billions, trillions, quadrillions and above, but for the purpose of this challenge, you don't have to handle any number above 999,999,999 (Nine Hundred and ninety-nine million, nine-hundred and ninety-nine thousand, nine hundred and ninety-nine.)

Python solver

Here is a short python script to verify answers:

import en def get_letter_num(s): count = 0 for c in s: if c.isalpha(): count += 1 return count number = input() count = 0 for i in xrange(1, number + 1): count += get_letter_num(en.number.spoken(i)) print count

Note that this usesthe NodeBox linguistics library to convert numbers to English. (yes, I just broke my own rule, but this isn't a competing answer) This is freely available here.

Sample I/O

7: 27 19: 106 72: 583 108: 1000 1337: 31,131 1234567: 63,448,174
Improve this question
\$\endgroup\$
13
  • 1
    \$\begingroup\$ Why is it One hundred and one, but then One million, six thousand, one hundred one without the and? \$\endgroup\$ Commented Feb 19, 2016 at 18:18
  • 1
    \$\begingroup\$ Related and more related. \$\endgroup\$ Commented Feb 19, 2016 at 18:32
  • 1
    \$\begingroup\$ @FryAmTheEggman Using his python script, 1100 -> one thousand and one hundred; 1200 -> one thousand two hundred, 1000100 -> one million and one hundred, 1000200 -> one millian two hundred. I think either A) DJ McGoathem should address the 1100 and 1000100 special cases in his question, or B) correct his test cases \$\endgroup\$ Commented Feb 24, 2016 at 18:00
  • 4
    \$\begingroup\$ Why the "and"? Proper names for numbers never use it: 123 = "one hundred twenty-three" \$\endgroup\$ Commented Feb 24, 2016 at 20:42
  • 1
    \$\begingroup\$ @ricdesi I agree. Related. People count "one thousand one, one thousand two, ...", without the ands. \$\endgroup\$ Commented Feb 24, 2016 at 21:49

1 Answer 1

Reset to default
1
\$\begingroup\$

Python 2, 266 259 236 229 228 bytes

This works for all inputs below one billion. This works for all test cases.

def l(n):b=[6,3,2][(n<1000)+(n<10**6)];c=10**b;return int("0335443554"[n%10])+int("0366555766"[n/10])+(n-10in[4,6,7,9])if n<100else l(n/c)+(l(n%c)or-3*(b<3))+7+(b<6)+2*(b<3)+3*(b>2)*(0<n%c<101) print sum(map(l,range(input()+1)))

To modify it to fit the question as stated (e.g. don't treat numbers ending with 100 special) simply replace the number 101 at the end of the first line with 100.

Explanation:

def l(n): b=[6, 3, 2][(n < 1000) + (n < 10**6)] # b = 2 if n < 1000 else 3 if n < 1000000 else 6 c=10**b return ( # Parenthesis added for readability. int("0335443554"[n % 10]) + # Compute length of last digit. one -> 3, seven -> 5, etc. int("0366555766"[n / 10]) + # Compute length of second to last digit. ten -> 3, eighty -> 6, etc. (n - 10 in[4, 6, 7, 9]) # Add one to length if the number is 14, 16, 17, or 19. if n < 100 else # Use above procedure if the number is under 100. # If otherwise, use below procedure. l(n / c) + # Compute length of the top portion of number. (l(n % c) or # Compute length of bottom portion of number. -3 * (b < 3)) + # If the number < 1000 and is a multiple of 100, # subtract 3 from the length because of missing and. 7 + # Add 7 to the length for "million" (b < 6) + # Add 8 to the length for "thousand" 2 * (b < 3) + # Add 10 to the length for "hundred and" 3 * # Add 3 to the length for another "and" (b > 2) * # if the number >= 1000 (0 < n % c < 101) # and the bottom portion > 0 and <= 100 ) print sum(map(l,range(input()+1))) # For the reader to figure out.
Improve this answer
\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.