Solution: Connecting n Pipes with Minimum Cost

int minCost(int pipes[], int n) {
  int cost;
  bool isFirstRun = true;
  int tempCost = 0;
  int tempPipes[n];
  std::sort(pipes, pipes + n); // built in function of a C++ library
  // find value for all combinations
  do {
    for (int pipeIndex = 0; pipeIndex < n; pipeIndex++) {
      tempPipes[pipeIndex] = pipes[pipeIndex];
    }
    for (int i = 0; i < n - 1; i++) {
      tempCost += (tempPipes[i] + tempPipes[i + 1]); //find current cost
      tempPipes[i + 1] = tempPipes[i] + tempPipes[i + 1];
    }
    if (isFirstRun) { //for first run add as it is in cost
      cost = tempCost;
      isFirstRun = false;
    } else if (tempCost < cost) //for later runs perform comparison
    {
      cost = tempCost;
    }
    tempCost = 0;
  } while (std::next_permutation(pipes, pipes + n)); // produces all combinations
  return cost;
}
int main() {
  int pipes[] = {4, 3, 2, 6 };
  int size = sizeof(pipes) / sizeof(pipes[0]);
  cout << "Total cost for connecting pipes list 1 is " << minCost(pipes, size);
  cout << endl << endl;
  int pipes1[] = {1, 1, 2, 6};
  int size1 = sizeof(pipes1) / sizeof(pipes1[0]);
  cout << "Total cost for connecting pipes list 2 is " << minCost(pipes1, size1);
  return 0;
}

int minCost(int pipes[], int n) {
  int cost;
  std::sort(pipes, pipes + n); // to generate all combinations.
   
    for (int i = 0; i < n - 1 ; i++) {
      int prevCost = cost; //store previous cost for later use
      
      cost = (pipes[i] + pipes[i+1]); //find current cost
      pipes[i + 1] = cost; //insert in array
      
      cost = cost + prevCost; //add with previous cost
    }
  
  return cost;
}
int main() {
  int pipes[] = {4, 3, 2, 6 };
  int size = sizeof(pipes) / sizeof(pipes[0]);
  cout << "Total cost for connecting pipes list 1 is " << minCost(pipes, size);
  cout << endl << endl;
  int pipes1[] = {1, 1, 2, 6};
  int size1 = sizeof(pipes1) / sizeof(pipes1[0]);
  cout << "Total cost for connecting pipes list 2 is " << minCost(pipes1, size1);
  return 0;
}