First Compass Challenge

From: 14/06/2023 to 21/06/2023

Instructions

• The challenge consists in five code challenges about Java, SQL and NoSQL
• Each code challenge must have its own production branch which will be merged to dev branch and then to main branch

Question	Jump to it
Question 1	Link
Question 2	Link
Question 3	Link
Question 4	Link
Question 5	Link

---

Code challenge 1

---

• Main language: JAVA

• Tasks
  • folder creation
  • branch creation
  • start code production
  • question done

About the challenge

Must output the student name that failed in the class respecting the quantity of problems solved and the name ordered alphabetically

First tiebraker => number of problems solved

Second tiebraker => last name alphabetically sorted

How to solve it

• Create one Array List to store the students name and another one to store how many problems were solved;
  • With an Array List will be easier remove data, add data and comparing it. The methods most used in the code.

  ArrayList<Integer> problemSolved = new ArrayList<Integer>();
  ArrayList<String> studentsName = new ArrayList<String>();

• Instantiate a Scanner Class to receive the data input;

   Scanner source = new Scanner(System.in);

  • With a loop that could be a for because we already know the number of students, each of them will be inserted in a different line, we will add:
    • Every single integer with the method .nextInt() (of the Scanner Class) to the problems Array List;
    • Every single string with the method .next() (of the Scanner Class) to the students name Array List;

    for (int i = 0; i < studentsQuantity; i++) {
    studentsName.add(source.next());
    problemSolved.add(source.nextInt());
    }

    • HERE COMES THE MAGIC, respecting the tiebreakers:
      • Using the j and i variables, the code will compare the values in the indexes that j and i points to

      for (int j = problemSolved.size() - 1, i = 0; j > 0; j--)

      • The first if statement will remove from students name Array List and from the problems solved Array List the value in the i index if it is greater than the value in the j index

      if (problemSolved.get(i) > problemSolved.get(j)) {
      problemSolved.remove(i);
      studentsName.remove(i);
      }

      • In line 22, if the value in the i index is lower than the value in the j index, the values in the j index will be removed from both Array Lists.

      else if (problemSolved.get(i) < problemSolved.get(j)) {
      problemSolved.remove(j);
      studentsName.remove(j);
      }

      • The last and most important, the piece of code below checks if both values (i and j) are equal to each other.

      else if (Objects.equals(problemSolved.get(i), problemSolved.get(j)))

      • If so, it means that both students got the same number of problems solved, and we will go for the second tiebraker;
      • The method str1.compareTo(str2) can return 3 distinct values, which can be:
        • An int value of 0 if the string is equal to the other string.
          • A case which will not happen because there are no homonyms

        else {
        System.out.println("Both students have the same name");
        break;
        }

        • An int value lower than 0 if the string is lexicographically less than the other string

        else if (studentsName.get(i).compareTo(studentsName.get(j)) < 0) {
        studentsName.remove(i);
        problemSolved.remove(i);
        }

        • An int value greater than 0 if the string is lexicographically greater than the other string (more characters)

        else if (studentsName.get(i).compareTo(studentsName.get(j)) < 0) {
        studentsName.remove(i);
        problemSolved.remove(i);
        }

        • At the end, the student with the least number os problems solved and with the last name alphabetically sorted will be printed out

        System.out.println(studentsName.get(0));

---

Code challenge 2

---

• Main language: JAVA

• Tasks
  • folder creation
  • branch creation
  • start code production
  • question done

About the challenge

Must output one of the next messages

• "Fun" => if the amount happy faces is greater than the amount of sad faces
• "Neutral" => if the amount of happy faces is equal than the amount of sad faces
• "Sad" => if the amount of happy faces is lower than the amount of happy faces

How to solve it

• Create an array (elements) to store each element from the string (line) separated by whitespaces using the method .split( );
• Instatiate a Scanner Class to read the input;
• With a string (line) store each .nextLine( )
• With two int variables

  Scanner source = new Scanner(System.in);
  String line = source.nextLine();
  String[] elements = line.split(" ");
  int upsetCount = 0, funCount = 0;

• A for each loop is used to count how many ":-(" and ":-)" are on the array (elements)

  for (String word : elements) {
      if (word.equals(":-(")){
          upsetCount++ ;
      } else if (word.equals(":-)")){
          funCount++;
      }
  }

• Finally, an if statement to check the numeric values of upsetCount and funCount;

  if (upsetCount==funCount){
      System.out.println("Neutral");
  } else if (upsetCount > funCount) {
      System.out.println("Upset");
  } else {
      System.out.println("Fun");
  }

---

Code challenge 3

---

• Main language: JAVA

• Tasks
  • folder creation
  • branch creation
  • start code production
  • question done

About the challenge

Must output the result of an equation

How to solve it

• Instantiate a Scanner Class to read the input;
• Create an array(results) to store each equation's result and then print it out

  Scanner source = new Scanner(System.in);
  int result = 0;
  List<Integer> results = new ArrayList<>();
  int k = 1;;

• A while loop to run through every line

  while (source.hasNext())

• Some if statement to check how many number are on the input and if it respects the constraints

  if (operandsQuantity < 1 || operandsQuantity > 100) {
  break;
  }
  if (operandsQuantity != 0) {
  String equation = source.next();

• This particular for loop was made to store the signs presented in the equationOperator.

  for (int i = 0; i < equationFormatted.length(); i++) {
        equationOperator.add(equationFormatted.charAt(i));
        }

• Another if statement to check other constraints related to the numbers of operands and how many of them were inserted

  if (operandsQuantity < equationCounter.size()) {
  System.out.println("Exceeded number of operands, you must have inserted " + operandsQuantity + " operands");
  
  } else if (operandsQuantity > equationCounter.size()) {
  System.out.println("You must have inserted " + operandsQuantity + " operands");

• The last part of the code was developed to add or to subtract the value of the result
• Then this value is stored in the array(results)

  } else{
  for(int j=0;j<equationCounter.size();j++){
  if(equationOperator.get(j).hashCode()=="-".hashCode()){

  result=result-Integer.parseInt((String)equationCounter.get(j));

  }else{
  result=Integer.parseInt((String)equationCounter.get(j))+result;
  }

  }
  results.add(result);
  }

• A for each loop to print each result stored in the array(results)

  for (int i : results) {
  System.out.println("Test: " + k);
  System.out.println(i);
  k++;

---

Code challenge 4

---

• Main language: SQL

• Tasks
  • folder creation
  • branch creation
  • start code production
  • question done

About the challenge

Must write a scritp for PostgreSQL creating in this order:

1. Create a database;
2. Create a table to store addresses with
   1. address_id
   2. zip code
   3. street name
   4. complement for the address
   5. neighborhood
   6. city name
   7. state name
   8. country name;
3. Create a table to store persons with
   1. person_id
   2. name
   3. age
   4. phone
   5. height
   6. email
   7. cpf
   8. date of birth
   9. address_id

Note: The address_id in Person Table should be a foreign key which refers to the address_id in the Address Table

How to solve it

• Using the SQL language create the first database;
• Then create the Address Table to first store the address to be referenced by the Person Table
• Finally, create the Person Table.

---

Code challenge 5

---

• Main language: NoSQL

• Tasks
  • folder creation
  • branch creation
  • start code production
  • question done

About the challenge

Must create a script for MondoDB inserting a Person Document containing:

• Object;
• Name;
• Age;
• Phone;
• Height;
• Email;
• CPF;
• Date of birth;
• Address > - Zip code;
  • Street name;
  • Complement;
  • Neighborhood;
  • City;
  • State;
  • Country

How to solve it

• Download Mongosh, you can find more about it here;
• Start the executable file in .zip folder;
• Connect to your MongoDB server;
• And then follow the written code.