<< 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 number), 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.

# My code

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

       #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;
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);

// 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.

# 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:

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)

# Benchmark

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

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

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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The 239 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
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.

 << 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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