<< problem 161 - Triominoes | Cross-hatched triangles - problem 163 >> |

# Problem 162: Hexadecimal numbers

(see projecteuler.net/problem=162)

In the hexadecimal number system numbers are represented using 16 different digits:

0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

The hexadecimal number AF when written in the decimal number system equals 10*16+15=175.

In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present.

Like numbers written in base ten we write hexadecimal numbers without leading zeroes.

How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once?

Give your answer as a hexadecimal number.

(A,B,C,D,E and F in upper case, without any leading or trailing code that marks the number as hexadecimal and without leading zeroes,

e.g. 1A3F and not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F)

# My Algorithm

A nice dynamic programming problem ... my function `count`

has 5 parameters:

- `digits`

stands for the number of digits, initially 16

- `haveAny`

is only true if any of the preceding digits is different from zero

- `haveZero`

is only true if at least one of the preceding digits is `0`

- `haveOne`

is only true if at least one of the preceding digits is `1`

- `haveA`

is only true if at least one of the preceding digits is `A`

When looking at the current digit then there are four groups:

1. anything except `0`

, `1`

, `A`

:

→ that means 13x everything with `digits - 1`

→ since it's not `0`

we have at least one digit different from zero, i.e. `haveAny`

must be true

2. current digit is `0`

:

→ if there was already a zero, it's the same like case 1

→ if there was no zero so far, then a zero is only allowed if `haveAny`

is true (no leading zero !)

3. current digit is `1`

:

→ if there was already a one, it's the same like case 1

→ if there was no `1`

so far, then set `haveOne`

to true, `haveAny`

as well

4. current digit is `A`

:

→ exactly the same thinking like case 3 but set `haveA`

to true instead of `haveOne`

I abort early if all conditions `haveZero`

, `haveOne`

and `haveA`

are fulfilled.

## Modifications by HackerRank

All results have to be modulo 10^9+7 and up to 100 digits are allowed.

The result should *not* be displayed in hexadecimal.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains `#ifdef`

s 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).

//#define ORIGINAL

#include <iostream>
#include <iomanip>
// recursively count solutions

unsigned long long count(unsigned int digits, bool haveAny = false,
bool haveZero = false, bool haveOne = false, bool haveA = false)
{
// solved ?
if (haveZero && haveOne && haveA && digits < 15)
return 1ULL << (4 * digits); // same as pow(16, digits);
// processed all digits ? (but no combination of 0, 1, A found)
if (digits == 0)
return 0;
// assume current digit is not 0, 1 or A
unsigned long long next = count(digits - 1, true, haveZero, haveOne, haveA);
unsigned long long result = 13 * next;
// try to use a zero (only allowed if already have any non-zero digit => "no leading zero")
result += haveZero ? next : count(digits - 1, haveAny, haveAny, haveOne, haveA);
// try to use 1
result += haveOne ? next : count(digits - 1, true, haveZero, true, haveA);
// try to use A
result += haveA ? next : count(digits - 1, true, haveZero, haveOne, true);
#ifndef ORIGINAL
result %= 1000000007ULL;
#endif
return result;
}
int main()
{
#ifdef ORIGINAL
std::cout << std::uppercase << std::hex << count(16) << std::endl;
#else
unsigned int digits;
std::cin >> digits;
std::cout << count(digits) << std::endl;
#endif
return 0;
}

This solution contains 8 empty lines, 7 comments and 7 preprocessor commands.

# 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.

This is equivalent to`echo 3 | ./162`

Output:

*Note:* the original problem's input `16`

__cannot__ be entered

because just copying results is a soft skill reserved for idiots.

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

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++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

May 25, 2017 submitted solution

May 25, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler162

My code solves **11** out of **11** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **45%** (out of 100%).

Hackerrank describes this problem as **medium**.

*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.

# Links

projecteuler.net/thread=162 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p134.java (written by Nayuki)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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

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I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

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

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort.

<< problem 161 - Triominoes | Cross-hatched triangles - problem 163 >> |