How can I remove the repetitive characters in Histogram [JAVA]










0















class Mclass {
public static void main(String args)

char a= 'a','b','c','d','a','b','c';
int count = 0;

for (int i=0; i<a.length; i++)

for(int j=0; j<a.length; j++)

if ( a[j] == a[i] )
count += 1;

System.out.println(a[i]+":"+count);
count = 0;




Output:



a:2
b:2
c:2
d:1
a:2
b:2
c:2



Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?










share|improve this question


























    0















    class Mclass {
    public static void main(String args)

    char a= 'a','b','c','d','a','b','c';
    int count = 0;

    for (int i=0; i<a.length; i++)

    for(int j=0; j<a.length; j++)

    if ( a[j] == a[i] )
    count += 1;

    System.out.println(a[i]+":"+count);
    count = 0;




    Output:



    a:2
    b:2
    c:2
    d:1
    a:2
    b:2
    c:2



    Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?










    share|improve this question
























      0












      0








      0








      class Mclass {
      public static void main(String args)

      char a= 'a','b','c','d','a','b','c';
      int count = 0;

      for (int i=0; i<a.length; i++)

      for(int j=0; j<a.length; j++)

      if ( a[j] == a[i] )
      count += 1;

      System.out.println(a[i]+":"+count);
      count = 0;




      Output:



      a:2
      b:2
      c:2
      d:1
      a:2
      b:2
      c:2



      Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?










      share|improve this question














      class Mclass {
      public static void main(String args)

      char a= 'a','b','c','d','a','b','c';
      int count = 0;

      for (int i=0; i<a.length; i++)

      for(int j=0; j<a.length; j++)

      if ( a[j] == a[i] )
      count += 1;

      System.out.println(a[i]+":"+count);
      count = 0;




      Output:



      a:2
      b:2
      c:2
      d:1
      a:2
      b:2
      c:2



      Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?







      java loops histogram






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 13 '18 at 19:49









      Haseeb AhmedHaseeb Ahmed

      175




      175






















          3 Answers
          3






          active

          oldest

          votes


















          0














          Starting from what you already did, first sort the array then try to count



          import java.util.*;
          class Mclass
          public static void main(String args)
          char a= 'a','b','c','d','a','b','c';
          int count = 0;
          Arrays.sort(a); // sort the array
          for (int i=0; i<a.length; i++)

          for(int j=i; j<a.length; j++)

          if ( a[j] == a[i] )
          count += 1;
          continue;

          i=j-1;
          break;

          System.out.println(a[i]+":"+count);
          count = 0;





          output



          a:2
          b:2
          c:2
          d:1





          share|improve this answer
































            1














            If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set and print only when Set doesn't contain the character and after printing, add it to Set so next time on wards it doesn't get printed.



            Change your code to this,



            class Mclass 
            public static void main(String args)
            Set<String> doneSet = new HashSet<String>();

            char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
            int count = 0;

            for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a.length; j++)
            if (a[j] == a[i])
            count += 1;

            if (!doneSet.contains(String.valueOf(a[i])))
            System.out.println(a[i] + ":" + count);
            doneSet.add(String.valueOf(a[i]));

            count = 0;





            This gives following output as you intend,



            a:2
            b:2
            c:2
            d:1





            share|improve this answer






























              0














              Do not print inside loop



              Save your count and print outside the loop.



              Do something like this:



              public class Mclass 
              public static void main(String args)

              char a= 'a','b','c','d','a','b','c';
              int count = 0;
              Map<String,Integer> output = new HashMap<>();

              for (int i=0; i<a.length; i++)

              for(int j=0; j<a.length; j++)

              if ( a[j] == a[i] )
              count += 1;

              output.put(Character.toString(a[i]), count);
              //System.out.println(a[i]+":"+count);
              count = 0;

              System.out.println(output);







              share|improve this answer






















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                0














                Starting from what you already did, first sort the array then try to count



                import java.util.*;
                class Mclass
                public static void main(String args)
                char a= 'a','b','c','d','a','b','c';
                int count = 0;
                Arrays.sort(a); // sort the array
                for (int i=0; i<a.length; i++)

                for(int j=i; j<a.length; j++)

                if ( a[j] == a[i] )
                count += 1;
                continue;

                i=j-1;
                break;

                System.out.println(a[i]+":"+count);
                count = 0;





                output



                a:2
                b:2
                c:2
                d:1





                share|improve this answer





























                  0














                  Starting from what you already did, first sort the array then try to count



                  import java.util.*;
                  class Mclass
                  public static void main(String args)
                  char a= 'a','b','c','d','a','b','c';
                  int count = 0;
                  Arrays.sort(a); // sort the array
                  for (int i=0; i<a.length; i++)

                  for(int j=i; j<a.length; j++)

                  if ( a[j] == a[i] )
                  count += 1;
                  continue;

                  i=j-1;
                  break;

                  System.out.println(a[i]+":"+count);
                  count = 0;





                  output



                  a:2
                  b:2
                  c:2
                  d:1





                  share|improve this answer



























                    0












                    0








                    0







                    Starting from what you already did, first sort the array then try to count



                    import java.util.*;
                    class Mclass
                    public static void main(String args)
                    char a= 'a','b','c','d','a','b','c';
                    int count = 0;
                    Arrays.sort(a); // sort the array
                    for (int i=0; i<a.length; i++)

                    for(int j=i; j<a.length; j++)

                    if ( a[j] == a[i] )
                    count += 1;
                    continue;

                    i=j-1;
                    break;

                    System.out.println(a[i]+":"+count);
                    count = 0;





                    output



                    a:2
                    b:2
                    c:2
                    d:1





                    share|improve this answer















                    Starting from what you already did, first sort the array then try to count



                    import java.util.*;
                    class Mclass
                    public static void main(String args)
                    char a= 'a','b','c','d','a','b','c';
                    int count = 0;
                    Arrays.sort(a); // sort the array
                    for (int i=0; i<a.length; i++)

                    for(int j=i; j<a.length; j++)

                    if ( a[j] == a[i] )
                    count += 1;
                    continue;

                    i=j-1;
                    break;

                    System.out.println(a[i]+":"+count);
                    count = 0;





                    output



                    a:2
                    b:2
                    c:2
                    d:1






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 13 '18 at 21:34

























                    answered Nov 13 '18 at 21:23









                    The Scientific MethodThe Scientific Method

                    1,7462515




                    1,7462515























                        1














                        If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set and print only when Set doesn't contain the character and after printing, add it to Set so next time on wards it doesn't get printed.



                        Change your code to this,



                        class Mclass 
                        public static void main(String args)
                        Set<String> doneSet = new HashSet<String>();

                        char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
                        int count = 0;

                        for (int i = 0; i < a.length; i++)
                        for (int j = 0; j < a.length; j++)
                        if (a[j] == a[i])
                        count += 1;

                        if (!doneSet.contains(String.valueOf(a[i])))
                        System.out.println(a[i] + ":" + count);
                        doneSet.add(String.valueOf(a[i]));

                        count = 0;





                        This gives following output as you intend,



                        a:2
                        b:2
                        c:2
                        d:1





                        share|improve this answer



























                          1














                          If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set and print only when Set doesn't contain the character and after printing, add it to Set so next time on wards it doesn't get printed.



                          Change your code to this,



                          class Mclass 
                          public static void main(String args)
                          Set<String> doneSet = new HashSet<String>();

                          char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
                          int count = 0;

                          for (int i = 0; i < a.length; i++)
                          for (int j = 0; j < a.length; j++)
                          if (a[j] == a[i])
                          count += 1;

                          if (!doneSet.contains(String.valueOf(a[i])))
                          System.out.println(a[i] + ":" + count);
                          doneSet.add(String.valueOf(a[i]));

                          count = 0;





                          This gives following output as you intend,



                          a:2
                          b:2
                          c:2
                          d:1





                          share|improve this answer

























                            1












                            1








                            1







                            If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set and print only when Set doesn't contain the character and after printing, add it to Set so next time on wards it doesn't get printed.



                            Change your code to this,



                            class Mclass 
                            public static void main(String args)
                            Set<String> doneSet = new HashSet<String>();

                            char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
                            int count = 0;

                            for (int i = 0; i < a.length; i++)
                            for (int j = 0; j < a.length; j++)
                            if (a[j] == a[i])
                            count += 1;

                            if (!doneSet.contains(String.valueOf(a[i])))
                            System.out.println(a[i] + ":" + count);
                            doneSet.add(String.valueOf(a[i]));

                            count = 0;





                            This gives following output as you intend,



                            a:2
                            b:2
                            c:2
                            d:1





                            share|improve this answer













                            If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set and print only when Set doesn't contain the character and after printing, add it to Set so next time on wards it doesn't get printed.



                            Change your code to this,



                            class Mclass 
                            public static void main(String args)
                            Set<String> doneSet = new HashSet<String>();

                            char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
                            int count = 0;

                            for (int i = 0; i < a.length; i++)
                            for (int j = 0; j < a.length; j++)
                            if (a[j] == a[i])
                            count += 1;

                            if (!doneSet.contains(String.valueOf(a[i])))
                            System.out.println(a[i] + ":" + count);
                            doneSet.add(String.valueOf(a[i]));

                            count = 0;





                            This gives following output as you intend,



                            a:2
                            b:2
                            c:2
                            d:1






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Nov 13 '18 at 20:06









                            Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi

                            8,0252927




                            8,0252927





















                                0














                                Do not print inside loop



                                Save your count and print outside the loop.



                                Do something like this:



                                public class Mclass 
                                public static void main(String args)

                                char a= 'a','b','c','d','a','b','c';
                                int count = 0;
                                Map<String,Integer> output = new HashMap<>();

                                for (int i=0; i<a.length; i++)

                                for(int j=0; j<a.length; j++)

                                if ( a[j] == a[i] )
                                count += 1;

                                output.put(Character.toString(a[i]), count);
                                //System.out.println(a[i]+":"+count);
                                count = 0;

                                System.out.println(output);







                                share|improve this answer



























                                  0














                                  Do not print inside loop



                                  Save your count and print outside the loop.



                                  Do something like this:



                                  public class Mclass 
                                  public static void main(String args)

                                  char a= 'a','b','c','d','a','b','c';
                                  int count = 0;
                                  Map<String,Integer> output = new HashMap<>();

                                  for (int i=0; i<a.length; i++)

                                  for(int j=0; j<a.length; j++)

                                  if ( a[j] == a[i] )
                                  count += 1;

                                  output.put(Character.toString(a[i]), count);
                                  //System.out.println(a[i]+":"+count);
                                  count = 0;

                                  System.out.println(output);







                                  share|improve this answer

























                                    0












                                    0








                                    0







                                    Do not print inside loop



                                    Save your count and print outside the loop.



                                    Do something like this:



                                    public class Mclass 
                                    public static void main(String args)

                                    char a= 'a','b','c','d','a','b','c';
                                    int count = 0;
                                    Map<String,Integer> output = new HashMap<>();

                                    for (int i=0; i<a.length; i++)

                                    for(int j=0; j<a.length; j++)

                                    if ( a[j] == a[i] )
                                    count += 1;

                                    output.put(Character.toString(a[i]), count);
                                    //System.out.println(a[i]+":"+count);
                                    count = 0;

                                    System.out.println(output);







                                    share|improve this answer













                                    Do not print inside loop



                                    Save your count and print outside the loop.



                                    Do something like this:



                                    public class Mclass 
                                    public static void main(String args)

                                    char a= 'a','b','c','d','a','b','c';
                                    int count = 0;
                                    Map<String,Integer> output = new HashMap<>();

                                    for (int i=0; i<a.length; i++)

                                    for(int j=0; j<a.length; j++)

                                    if ( a[j] == a[i] )
                                    count += 1;

                                    output.put(Character.toString(a[i]), count);
                                    //System.out.println(a[i]+":"+count);
                                    count = 0;

                                    System.out.println(output);








                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Nov 13 '18 at 20:02









                                    Sayantan MandalSayantan Mandal

                                    273213




                                    273213



























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