My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, as well as the input data, 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 <set>
       
#include <map>
       
#include <string>
       
#include <iostream>
       
 
       
int main()
       
{
       
  unsigned int logins = 50; // p079_keylog.txt contains 50 logins
       
 
       
//#define ORIGINAL

       
#ifndef ORIGINAL
       
  std::cin >> logins;
       
#endif
       
 
       
  // read all logged inputs
       
  // for each digit/letter, store its predecessor
       
  std::map<char, std::set<char>> previous;
       
  while (logins--)
       
  {
       
    std::string line;
       
    std::cin >> line;
       
    // create an empty set for the initial letter (if it doesn't exist yet)
       
    previous[line[0]];
       
    // and for the other letters, store their predecessors
       
    for (unsigned int i = 1; i < line.size(); i++)
       
      previous[line[i]].insert(line[i - 1]);
       
  }
       
 
       
  // until we have no characters left ...
       
  std::string result;
       
  while (!previous.empty())
       
  {
       
    // find lexicographically smallest letter with no predecessor
       
    auto emptySet = previous.begin();
       
    while (emptySet != previous.end() && !emptySet->second.empty())
       
      emptySet++;
       
 
       
    // invalid ?
       
    if (emptySet == previous.end())
       
    {
       
      result = "SMTH WRONG"; // Hackerrank's message if code cannot be decrypted
       
      break;
       
    }
       
 
       
    // print letter
       
    auto current = emptySet->first;
       
    result += current;
       
 
       
    // that letter won't appear in the keyphrase anymore
       
    previous.erase(current);
       
 
       
    // remove from the predecessor list of all other letters
       
    for (auto& p : previous)
       
      p.second.erase(current);
       
  }
       
 
       
  // print keyphrase
       
  std::cout << result << std::endl;
       
  return 0;
       
}

This solution contains 10 empty lines, 12 comments and 6 preprocessor commands.