<< problem 16 - Power digit sum Maximum path sum I - problem 18 >>

# Problem 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.

# My Algorithm

My program converts a number into its "written" representation because of the Hackerrank requirements (see below).
A simple loop from 1 to 1000 creates a ton of string and computes the sums their lengths.

The function convert immediately returns the name of numbers in [0;19].
For all other numbers it calls itself recursively:
e.g. when the parameter x is in [20;99] then its higher digit is converted directly into a word, its lower is found by a recursive call

My code is a bit bloated because of spelling differences between Project Euler and Hackerrank.
I had to be a bit careful not to call the function convert recursively with parameter zero.

## Alternative Approaches

The original problem can be solved by just counting the letter without actually "building" the names, too.

## Modifications by HackerRank

The Hackerrank problem is quite different: you have to convert a number into its name.
Their spelling rules vary, too.

Note: Unlike most of my other programs, #define ORIGINAL is not active in the source code listing due to interactive tests.

## Note

Rules for finding the English names of numbers have far less exceptions than the German rules ...

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 17" | ./17

Output:

(please click 'Go !')

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <string>

// customize code for original problem
//#define ORIGINAL

// convert a number into its written representation
std::string convert(unsigned long long x)
{
#ifdef ORIGINAL
const std::string Gap = " And "; // British
const std::string ConnectTensAndOnes = "-";
#else
const std::string Gap = " ";
const std::string ConnectTensAndOnes = " ";
#endif

// none-composite names
switch(x)
{
case  0: return "Zero";
case  1: return "One";
case  2: return "Two";
case  3: return "Three";
case  4: return "Four";
case  5: return "Five";
case  6: return "Six";
case  7: return "Seven";
case  8: return "Eight";
case  9: return "Nine";
case 10: return "Ten";
case 11: return "Eleven";
case 12: return "Twelve";
case 13: return "Thirteen";
case 14: return "Fourteen";
case 15: return "Fifteen";
case 16: return "Sixteen";
case 17: return "Seventeen";
case 18: return "Eighteen";
case 19: return "Nineteen";
default: break;
}

// two-digit composite names
if (x >= 20 && x < 100)
{
auto ones = x % 10;
auto tens = x / 10;
auto strOnes = (ones != 0) ? ConnectTensAndOnes + convert(ones) : "";
switch (tens)
{
case 2: return "Twenty"  + strOnes;
case 3: return "Thirty"  + strOnes;
case 4: return "Forty"   + strOnes; // <= often misspelt/misspelled ;)
case 5: return "Fifty"   + strOnes;
case 6: return "Sixty"   + strOnes;
case 7: return "Seventy" + strOnes;
case 8: return "Eighty"  + strOnes;
case 9: return "Ninety"  + strOnes;
default: break; // never reached
}
}

// three-digit composite names
if (x >= 100 && x < 1000)
{
auto onesAndTens = x % 100;
auto hundreds    = x / 100;
auto strOnesAndTens = (onesAndTens != 0) ? Gap + convert(onesAndTens) : "";
return convert(hundreds) + " Hundred" + strOnesAndTens;
}

// four to six digits
if (x >= 1000 && x < 1000000)
{
auto low  = x % 1000; // variable names got too long, I went for a generic one ...
auto high = x / 1000;
auto strLow = (low != 0) ? Gap + convert(low) : "";
return convert(high) + " Thousand" + strLow;
}

// seven to nine digits
if (x >= 1000000 && x < 1000000000)
{
auto low  = x % 1000000;
auto high = x / 1000000;
auto strLow = (low != 0) ? Gap + convert(low) : "";
return convert(high) + " Million" + strLow;
}

// ten to twelve digits
if (x >= 1000000000 && x < 1000000000000ULL) // careful: must be a 64 bit constant, add "LL"
{
auto low  = x % 1000000000;
auto high = x / 1000000000;
auto strLow = (low != 0) ? Gap + convert(low) : "";
return convert(high) + " Billion" + strLow;
}

// thirteen to fifteen digits
if (x >= 1000000000000ULL && x < 1000000000000000ULL)
{
auto low  = x % 1000000000000ULL;
auto high = x / 1000000000000ULL;
auto strLow = (low != 0) ? Gap + convert(low) : "";
return convert(high) + " Trillion" + strLow;
}

// not reached
return "?";
}

int main()
{
#ifdef ORIGINAL
// count number of letters
unsigned int sum = 0;
for (unsigned int i = 1; i <= 1000; i++)
{
auto name = convert(i);
for (auto c : name)
if (std::isalpha(c)) // discard spaces/hyphens/etc.
sum++;
}
std::cout << sum << std::endl;
#else
// just print several names according to input
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned long long x;
std::cin >> x;
std::cout << convert(x) << std::endl;
}
#endif
return 0;
}


This solution contains 11 empty lines, 13 comments and 8 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 24, 2017 submitted solution

# Hackerrank

My code solves 6 out of 6 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 16 - Power digit sum Maximum path sum I - problem 18 >>
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