<< problem 169 - Exploring the number of different ways a number ... Finding numbers for which the sum of the squares ... - problem 171 >>

# Problem 170: Find the largest 0 to 9 pandigital that can be formed by concatenating products

Take the number 6 and multiply it by each of 1273 and 9854:
6 * 1273 = 7638
6 * 9854 = 59124

By concatenating these products we get the 1 to 9 pandigital 763859124. We will call 763859124 the "concatenated product of 6 and (1273,9854)". Notice too, that the concatenation of the input numbers, 612739854, is also 1 to 9 pandigital.

The same can be done for 0 to 9 pandigital numbers.

What is the largest 0 to 9 pandigital 10-digit concatenated product of an integer with two or more other integers, such that the concatenation of the input numbers is also a 0 to 9 pandigital 10-digit number?

# My Algorithm

I analyze all pandigital numbers, beginning with the largest (9876543210) and slowly decrement towards 1023456789.
The easiest way is to store the pandigital number in a std::string and call std::prev_permutation repeatedly (see my variable current).

A bold assumption is that the pandigital number only needs to be split into two parts left and right such that
left = factor * one and right = factor * two. If concat(factor, one, two) is 10-pandigital then we found the result.

Neither left nor right can begin with a zero. Any potential factor must be a divisor of left and right: 1 < factor <= gcd(left, right)
When looking at all 10-pandigital numbers (there are only 10! = 3628800 such numbers), I observed that all factor are multiples of 3.

## Modifications by HackerRank

The program is forced to use a certain output format (no big problem).
My program fails to process the massive amount of test cases.

## Note

Chances are that splitting the pandigital number into more than two parts produces valid results, too.
Maybe I just got lucky by finding the correct solution.

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 1987654320" | ./170

Output:

(please click 'Go !')

Note: the original problem's input 9876543210 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.

       #include <string>
#include <iostream>
#include <algorithm>

// true, if x contains only distinct digits
bool isPandigital(unsigned long long x)
{
unsigned char used[10] = { 0 };
while (x > 0)
{
auto digit = x % 10;
// digit already used ?
if (used[digit] == 1)
return false;
used[digit]++;
x /= 10;
}

return true;
}

// greatest common divisor
unsigned int gcd(unsigned int a, unsigned int b)
{
while (a != 0)
{
unsigned int c = a;
a = b % a;
b = c;
}
return b;
}

int main()
{
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
std::string current = "9876543210";
std::cin >> current;

// find next smaller pandigital number (only to avoid malicious input)
unsigned long long adjusted = std::stoll(current);
if (adjusted < 1023456789)

// start search
bool found = false;
do
{
// split into two parts and check each common divisor
for (size_t split = 1; split < current.size() && !found; split++)
{
// must not begin with a zero
if (current[0] == '0' || current[split] == '0')
continue;

auto left  = std::stoll(current.substr(0, split));
auto right = std::stoll(current.substr(split));

// any common divisors ?
unsigned int shared = gcd(left, right);
const unsigned int MultipleOfThree = 3; // I saw that all divisors are always multiples of three
for (unsigned int factor = MultipleOfThree; factor <= shared; factor += MultipleOfThree)
{
// analyze all common divisors
if (left  % factor == 0 &&
right % factor == 0)
{
// combine all digits
unsigned int one = left  / factor;
unsigned int two = right / factor;
std::string sequence = std::to_string(factor) +
std::to_string(one) +
std::to_string(two);

// must have exactly 10 pandigital digits
if (sequence.size() == 10 && isPandigital(std::stoll(sequence)))
{
found = true;
std::cout << factor << "*(" << one << "," << two << ")=" << current << std::endl;
break;
}
}
}
}

// done ?
if (found)
break;
} while (std::prev_permutation(current.begin(), current.end()));
}

return 0;
}


This solution contains 11 empty lines, 12 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.08 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

June 19, 2017 submitted solution

# Hackerrank

My code solves 2 out of 21 test cases (score: 5%)

I failed 0 test cases due to wrong answers and 19 because of timeouts

# Difficulty

Project Euler ranks this problem at 70% (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 169 - Exploring the number of different ways a number ... Finding numbers for which the sum of the squares ... - problem 171 >>
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