<< problem 110 - Diophantine reciprocals II Bouncy numbers - problem 112 >>

# Problem 111: Primes with runs

Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same:
1111 is divisible by 11, 2222 is divisible by 22, and so on.
But there are nine 4-digit primes containing three ones:
1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111

We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit,
N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.

So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit,
there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275.
It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.

In the same way we obtain the following results for 4-digit primes.

Digit dM(4,d)N(4,d)S(4,d)
021367061
13922275
2312221
331246214
4328888
5315557
6316661
73957863
8318887
93748073

For d = 0 to 9, the sum of all S(4, d) is 273700.

Find the sum of all S(10, d).

# My Algorithm

search iterates through all numbers and calls isPrime to figure out whether it's a prime number.
There are a few gotchas:
1. While iterating, I create all possible strings. They may contain leading zeros, which are not allowed.
2. Add numbers into the sequence of repeated digits only if the number's digits are in non-decreasing order: 125 is okay, 122 is okay, 121 isn't.

## Modifications by HackerRank

I should analyze numbers with up to 40 digits (instead of 10). My code is too slow to handle anything with more than 12 digits.
Even worse, my routines can't process numbers exceeding 64 bits (which is about 19 digits).

# Interactive test

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

Input data (separated by spaces or newlines):
Note: Number of digits

This is equivalent to
echo 4 | ./111

Output:

Note: the original problem's input 10 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.

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 <vector>
#include <algorithm>

#define ORIGINAL

// return true if x is prime, based on trial division (not suitable for large numbers)
bool isPrime(unsigned long long x)
{
// trial divison by 2
if (x % 2 == 0)
return x == 2;

// trial division by odd numbers
for (unsigned long long i = 3; i*i <= x; i++)
if (x % i == 0)
return false;

return true;
}

// find all numbers where "digit" is repeated "repeat" times and "extraDigits" digits are added
// e.g. 10007 is found by search(0, 3, 2)
unsigned long long search(unsigned int digit, unsigned int repeat, unsigned int extraDigits, bool printPrimes = false)
{
// sum of all matching primes
unsigned long long sum = 0;
std::vector<unsigned long long> matches; // Hackerrank only: print matches in ascending order

// create string with a digit repeated a few times
std::string sameDigit(repeat, digit + '0');

// insert numbers i where 0 <= i < maxExtra
// e.g. if extraDigits = 3 then from 0 to 999 (actually from "000" to "999")
unsigned long long maxExtra = 1;
for (unsigned int i = 1; i <= extraDigits; i++)
maxExtra *= 10;

// let's insert all those numbers !
for (unsigned long long extra = 0; extra < maxExtra; extra++)
{
// convert to string
auto current = std::to_string(extra);
// is in ascending order ?
auto sorted = current;
std::sort(sorted.begin(), sorted.end());
if (current != sorted)
continue;

// add zeros if "extra" has too few digits
current = '0' + current;

// and add the bunch of identical digits
current += sameDigit;
// sort everything to ensure that std::next_permutation works properly
std::sort(current.begin(), current.end());

do
{
if (current.front() == '0')
continue;
// no even numbers
if (current.back() % 2 == 0) // ASCII code of even digits is even, too
continue;

// is it a prime ?
unsigned long long num = std::stoll(current);
if (isPrime(num))
{
sum += num;
if (printPrimes)
matches.push_back(num);
}
} while (std::next_permutation(current.begin(), current.end()));
}

// Hackerrank only: print all primes we found
if (printPrimes && !matches.empty())
{
std::sort(matches.begin(), matches.end());
for (auto x : matches)
std::cout << x << " ";
}

return sum;
}

int main()
{
#ifdef ORIGINAL
// total number of digits
unsigned int digits = 10;
std::cin >> digits;
// sum of all matching primes
unsigned long long sum = 0;

// iterate over all repeated digits
for (unsigned int i = 0; i <= 9; i++)
{
// try to use as many repeated digits as possible
for (unsigned int repeated = digits - 1; repeated >= 1; repeated--)
{
auto insertDigits = digits - repeated;
auto found = search(i, repeated, insertDigits);
// found at least one prime ?
if (found > 0)
{
sum += found;
break;
}
}
}

std::cout << sum << std::endl;

#else

unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int digits;
unsigned int sameDigit;
std::cin >> digits >> sameDigit;

if (digits < 19) // my code can't handle > 64 bits at the moment
{
// iterate over all repeated digits
for (unsigned int repeated = digits - 1; repeated >= 1; repeated--)
{
auto insertDigits = digits - repeated;
if (search(sameDigit, repeated, insertDigits, true) > 0)
break;
}
}

std::cout << std::endl;
}
#endif

return 0;
}


This solution contains 23 empty lines, 25 comments and 8 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

May 17, 2017 submitted solution

# Hackerrank

My code solves 1 out of 20 test cases (score: 0%)

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

# Difficulty

Project Euler ranks this problem at 45% (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
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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.

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