<< problem 23 - Non-abundant sums 1000-digit Fibonacci number - problem 25 >>

# Problem 24: Lexicographic permutations

A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.

The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

# My Algorithm

The original problem can be solved with STL's handy std::next_permutation.
Even though we call it 999999 times it is still very fast (< 10 ms).

Note: the first permutation has index 0, hence 999999 instead of 1000000 iterations.

## Alternative Approaches

Hackerrank's modified problem has a much larger "search space" and causes timeout.
Therefore I implemented an alternative algorithm based on the "Factorial number system" (see en.wikipedia.org/wiki/Factorial_number_system):

499999 in our decimal system is:
4 * 10^4 + 9 * 10^4 + 9 * 10^3 + 9 * 10^2 + 9 * 10^1 + 9 * 10^0

The factorial number system replaces 10^x by x!:
1 * 9! + 3 * 8! + 3 * 7! + 1 * 6! + 2 * 5! + 3 * 4! + 1 * 3! + 0 * 2! + 1 * 1! + 0 * 0!
= 1 * 362880 + 3 * 40320 + 3 * 5040 + 1 * 720 + 2 * 120 + 3 * 24 + 1 * 6 + 0 * 2 + 1 * 1 + 0 * 1
= 499999

The coefficients 1, 3, 3, 1, 2, 3, 1, 0, 1, 0 define which indices of our original, unpermutated string we have to choose.
But there is a twist: whenever we select an element, we have to remove it from the original. And everything's 0-based.
That means:
0123456789 → choose element 1: 11
023456789  → choose element 3: 414
02356789  → choose element 3: 5145
0236789  → choose element 1: 21452
036789  → choose element 2: 614526
03789  → choose element 3: 8145268
0379  → choose element 1: 31452683
079  → choose element 0: 014526830
79  → choose element 1: 9145268309
7  → choose element 0: 71452683097

## Modifications by HackerRank

The string abcdefghijklm is used instead of 0123456789.
In the end, we have 13! instead of 10! potential permutations (1715x more permutations).

## Note

The "live test" is based on the Hackerrank problem.

# 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 500000" | ./24

Output:

(please click 'Go !')

Note: the original problem's input 1000000 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)

# 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>
#include <algorithm>

int main()
{
//#define ORIGINAL
#ifdef ORIGINAL
unsigned int numPermutation = 1000000;
std::string current = "0123456789";
while (--numPermutation)
std::next_permutation(current.begin(), current.end());
std::cout << current << std::endl;

#else

const std::string abc = "abcdefghijklm";
unsigned int tests;
std::cin >> tests;
while (tests--)
{
// to find the permutation we treat the input number as something written
// in a "factorial" system:
// x = pos0 * 12! + pos1 * 11! + pos2 * 10! + ... + pos12 * 1!
// (we have 13 letters, therefore the position of the first letter is in 12! radix)

// precomputed 0! .. 12!
const unsigned long long factorials[13+1] =
{ 1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800 };

// 13! which exceed 32 bits
unsigned long long x;
std::cin >> x;

// reduce to a single cycle (repeats after 13! iterations)
x %= factorials[abc.size()];

// that factorial system is zero-based ...
x--;

// strip off single letters (until empty)
auto remain = abc;
// our wanted permutation
std::string result;
while (!remain.empty())
{
// get next digit in that strange number system :-)
auto currentFactorial = factorials[remain.size() - 1];
auto pos = x / currentFactorial;

// store the associated letter
result += remain[pos];
// and remove it from the still unprocessed data
remain.erase(pos, 1);

// eliminate the processed digit
x %= currentFactorial;
}

std::cout << result << std::endl;
}
#endif
return 0;
}


This solution contains 11 empty lines, 15 comments and 6 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 23, 2017 submitted solution
April 4, 2017 added comments

# Hackerrank

My code solves 10 out of 10 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 23 - Non-abundant sums 1000-digit Fibonacci number - problem 25 >>
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