# Project Euler 17: Number letter counts

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.

## Solution

Numbers between 0 and 12 have unique names, from 13 to 19 the names have a teen-suffix and numbers between 20 and 99 are constructed by prefixing a tenth-prefix, except for zero, there is no twenty-zero, but of course a twenty-one and twenty-two. These ten-prefixes get suffixes between one and nine. This information can already be coded:

// proper nouns var proper = [ 0, "one".length, "two".length, "three".length, "four".length, "five".length, "six".length, "seven".length, "eight".length, "nine".length, "ten".length, "eleven".length, "twelve".length, "thirteen".length, "fourteen".length, "fifteen".length, "sixteen".length, "seventeen".length, "eighteen".length, "nineteen".length ]; // tenth prefixes var tenth = [ "twenty".length, "thirty".length, "forty".length, "fifty".length, "sixty".length, "seventy".length, "eighty".length, "ninety".length ]; // Returns the length of the numbers between 0 and 99 function below100(n) { if (n < 20) return proper[n]; return tenth[n / 10 - 2 | 0] + proper[n % 10]; }

So far so good. We can now calculate the length of the numbers between 0 and 99. After that, every number below 1000 gets a hundred suffix, like 234 is two-hundred plus the same as a number below 100. The question reminds us, that the tenths are added with an "and", like two-hundred and thirty-four, but only if it's not exactly a hundreds, like three-hundred. If a number is greater than 999, it's only necessary to append the thousandth part. Implemented, this can look like:

function numberLength(n) { if (n < 100) return below100(n); var res = 0; var h = Math.floor(n / 100) % 10; var t = Math.floor(n / 1000); var s = n % 100; if (n > 999) res+= below100(t) + "thousand".length; if (h !== 0) res+= proper[h] + "hundred".length; if (s !== 0) res+= "and".length + below100(s); return res; }

Thats cool we have a \(O(1)\) solution to calculate the length of a number name, which can now be used to loop over 1000 times to come up with the final solution:

function solution(n) { var num = 0; for (var i = 1; i <= n; i++) { num+= numberLength(i); } return num; } solution(1000);