## The challenge

You probably know that some characters written on a piece of paper, after turning this sheet 180 degrees, can be read, although sometimes in a different way. So, uppercase letters “H”, “I”, “N”, “O”, “S”, “X”, “Z” after rotation are not changed, the letter “M” becomes a “W”, and Vice versa, the letter “W” becomes a “M”.

We will call a word “shifter” if it consists only of letters “H”, “I”, “N”, “O”, “S”, “X”, “Z”, “M” and “W”. After turning the sheet, this word can also be read, although in a different way. So, the word “WOW “turns into the word “MOM”. On the other hand, the word “HOME” is not a shifter.

Find the number of unique shifter words in the input string (without duplicates). All shifters to be counted, even if they are paired (like “MOM” and “WOW”). String contains only uppercase letters.

#### Examples

 `````` 1 2 3 4 5 6 7 8 9 10 11 `````` ``````// shifter words are "SOS" and "IN" Shifter.count("SOS IN THE HOME") == 2 // shifter words are "WHO", "IS", "NO" Shifter.count("WHO IS SHIFTER AND WHO IS NO") == 3 // no shifter words Shifter.count("TASK") == 0 // no shifter words in empty string Shifter.count("") == 0 ``````

## The solution in Java code

Option 1:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 `````` ``````import java.util.*; public class Shifter{ public static int count(String st){ Map m = new HashMap<>(); for (String word : st.split("\\s+")) { if (isShifter(word)) m.put(word, 1); } return m.size(); } public static boolean isShifter(String word) { if (word==null || word=="") return false; for (char c : word.toCharArray()) { if ("HINOSXZMW".indexOf(c)==-1) return false; } return true; } } ``````

Option 2:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 `````` ``````import java.util.Arrays; import java.util.LinkedHashSet; public class Shifter{ public static int count(String st){ String[] stringArray = st.split(" "); LinkedHashSet noduplicate = new LinkedHashSet<>(Arrays.asList(stringArray)); stringArray = noduplicate.toArray(new String[]{}); int count = 0; for(String s: stringArray){ if(s.matches("[HINOSXZMW]+"))count++; } return count; } } ``````

Option 3:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 `````` ``````import java.util.Arrays; import java.util.regex.Pattern; public class Shifter{ public static int count(String value){ Pattern pattern = Pattern.compile("^[HINOSXZMW]+\$"); return (int) Arrays.stream(value.split("\\s+")) .distinct() .filter(token -> pattern.matcher(token).find()) .count(); } } ``````

## Test cases to validate our solution

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 `````` ``````import org.junit.Test; import static org.junit.Assert.assertEquals; import org.junit.runners.JUnit4; public class SolutionTest{ @Test public void on(){ assertEquals(1, Shifter.count("ON")); } @Test public void osIsUpdated(){ assertEquals(2, Shifter.count("OS IS UPDATED")); } @Test public void whoIsWho(){ assertEquals(2, Shifter.count("WHO IS WHO")); } @Test public void js(){ assertEquals(0, Shifter.count("JS")); } @Test public void iIiiIIii(){ assertEquals(2, Shifter.count("I III I III")); } @Test public void empty(){ assertEquals(0, Shifter.count("")); } } ``````