how to compare the letters of a String that are written of distinctly

For what you ask, that different cases of the same run can be evaluated, the following would be done:

Let's say the following case:

    public String orderChars(String _str)
    {
        char[] arrStr = _str.toCharArray();
        Arrays.sort(arrStr);
        StringBuilder str = new StringBuilder();
        int i;
        for (i = 0; i < arrStr.length; i++)
            str.append(arrStr[i]);

        return str.toString();
    }

    int casos = 4;
    String palabrasTrifelias[] = {"monja", "jamon", "copa" ,"paco" ,"pavio" ,"viona", "masa", "mesa"};

    Boolean resultado[] = new Boolean[casos];  //Aquí vamos a almacenar si están bien o no. Por defecto se inicializa todo a false, lo tenemos que pasar todo a true para la comparación que haremos luego.

    for(int i=0; i<casos; i++)
    {
        resultado[i] = true;
    }

    //Vamos mirando todas las palabras de 2 en 2
    for(int i=0; i< palabrasTrifelias.length; i=i+2)
    {
        String p1 = palabrasTrifelias[i];
        p1 = p1.replace("b", "v"); //Sustituimos las posibles b por v
        String p2 = palabrasTrifelias[i+1];
        p2 = p2.replace("b", "v");

        //Ahora ya podemos comparar las palabras de 2 en 2.
        //Primero tienen que medir lo mismo, sino ya no son iguales
        //Tampoco puede ser la misma palabra
        if(p1.length()==p2.length() && !p1.equals(p2))
        {
            p1 = orderChars(p1);  //Ordenamos los caracteres
            p2 = orderChars(p2);  //Ordenamos los caracteres

            for(int j = 0; j<p1.length(); j++)
            {
                if(p1.charAt(j)!=p2.charAt(j)){resultado[i/2]=false;} //Como están ordenados, comparamos 1 a 1 todos los caracteres
            }
        }else{
            resultado[i/2]=false;
        }
    }

//Ahora imprimimos los resultados
    System.out.println("Output: ");
    for(int i=0; i<casos; i++)
    {
        System.out.println(resultado[i]+ " ");
    }

  

You could do it this way too, as I mentioned in the comment with charAt ()

public static void main(String[] args) {

    Scanner Entrada = new Scanner(System.in);
    int Casos;

    System.out.print("Ingrese cantidad de casos: ");
    Casos = Entrada.nextInt();
    Casos = Casos*2;

    String PalabrasTrifelias[] = new String[Casos];

    for( int i = 0; i < PalabrasTrifelias.length; i++ ){
        PalabrasTrifelias[i] = Entrada.next();
        if( i % 2 == 0){
            System.out.println();
        }
    }

    int nx = 0;
    for ( String Palabras:PalabrasTrifelias){
        System.out.printf(" %d : %s%n", nx, Palabras);
        nx++;
    }

    nx = 0;
    for ( String Palabras:PalabrasTrifelias){
        String n1 = Palabras;
        nx++;
        String n2 = Palabras;
        int contador = 0;
        // Verifico que tengan el mismo numero de carateres creo que deberia tener el mismo numero
        if( n1.length() == n2.length()){
            // Ciclo para recorrer el string n1
            for(int i = 0; i < n1.length(); i++){
                // Ciclo para recorrer string n2
                for (int j = 0; j < n2.length(); j++){
                    // condicinal para ir comparando
                    if( n1.charAt(i) == n2.charAt(j)){
                        contador++;
                    }
                }
            }
        } 
        if (contador == n1.length()){
            System.out.println("Si cumple"+ n1 + " con "+ n2);
        }else{
            System.out.println("No cumple"+ n1 + " con "+ n2);
        }
        nx++;
    }
} 

Another possible implementation:

public static void main(String[] args) {
    trifelio("monja", "jamon");
    trifelio("Paco", "copa");
    trifelio("carro", "roca");
    trifelio("lavese", "besela");
    trifelio("vota", "bota");
}

private static void trifelio(String str1, String str2) {
    System.out.println("esTrifelio(" + str1 + ", " + str2 + "): " + esTrifelio(str1, str2));
}

private static boolean esTrifelio(String str1, String str2) {
    if(str1 == str2) {
        return false;
    }

    str1 = str1.replace("v", "b").toLowerCase();
    str2 = str2.replace("v", "b").toLowerCase();

    if(str1.equals(str2)) {
        return false;
    }

    return str1.concat(str1).contains(str2) && str2.concat(str2).contains(str1);
}

The idea behind is that if 2 Strings A, B are triplets, then you have to:

• AA contains B
• BB contains A

You can pose it as long as the two words have the same length, of course if one has more letters than the other obviously would no longer be triplet words, what I did was to transform the strings into arrays of characters and compare in this the first position of the first word, that is, the m of a nun with each of the letters of the second word, if it was found, it added one to the variable of integer type called coincidence, at the end of each position of the first word questions if what is worth coinciding is equal to the length of your string, if so, they are triplet words, otherwise, they are not triplet words

String palabra1 = "monja";
String palabra2 = "jamon";
int coincidencia = 0;

char[] Caracteres1 = palabra1.toCharArray();
char[] Caracteres2 = palabra2.toCharArray();

for (int x=0;x<Caracteres1.length;x++){
    for (int y=0;y<Caracteres2.length;y++){
      if(Caracteres1[x] == Caracteres2[y]){
          coincidencia++;
      }
    }
}

if(Caracteres1.length == coincidencia){
    System.out.println("SI");
}
else{
    System.out.println("NO");
}

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