Find duplicate characters in a String and count the number of occurances using Java

Question

How can I find the number of occurrences of a character in a string?

For example: The quick brown fox jumped over the lazy dog.

Some example

Answer 1

public static void main(String args[]) {
    char Char;
    int count;
    String a = "Hi my name is Rahul";
    a = a.toLowerCase();
    for (Char = 'a'; Char <= 'z'; Char++) {
        count = 0;
        for (int i = 0; i < a.length(); i++) {
            if (a.charAt(i) == Char) {
                count++;
            }
        }
        System.out.println("Number of occurences of " + Char + " is " + count);
    }
}

Answer 2

import java.io.*;
public class CountChar 
{

    public static void main(String[] args) throws IOException
    {
      String ch;
      BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
      System.out.print("Enter the Statement:");
      ch=br.readLine();
      int count=0,len=0;
        do
        {  
          try
          {
          char name[]=ch.toCharArray();
              len=name.length;
              count=0;
              for(int j=0;j<len;j++)
               {
                  if((name[0]==name[j])&&((name[0]>=65&&name[0]<=91)||(name[0]>=97&&name[0]<=123))) 
                      count++;
               }
              if(count!=0)
                System.out.println(name[0]+" "+count+" Times");
              ch=ch.replace(""+name[0],"");          
          }
          catch(Exception ex){}
        }
        while(len!=1);
   }

}

Output

Enter the Statement:asdf23123sfsdf

a 1 Times

s 3 Times

d 2 Times

f 3 Times

Answer 3

If your string only contains alphabets then you can use some thing like this.

public class StringExample {
public static void main(String[] args) {
    String str = "abcdabghplhhnfl".toLowerCase();
    // create a integer array for 26 alphabets.
    // where index 0,1,2.. will be the container for frequency of a,b,c...
    Integer[] ar = new Integer[26];
    // fill the integer array with character frequency.
    for(int i=0;i<str.length();i++) {
        int j = str.charAt(i) -'a';
        if(ar[j]==null) {
            ar[j]= 1;
        }else {
            ar[j]+= 1;
        }
    }   
    // print only those alphabets having frequency greater then 1.
    for(int i=0;i<ar.length;i++) {
        if(ar[i]!=null && ar[i]>1) {
            char c = (char) (97+i);
            System.out.println("'"+c+"' comes "+ar[i]+" times.");
        }
    }
}
}

Output:

 'a' comes 2 times.
 'b' comes 2 times.
 'h' comes 3 times.
 'l' comes 2 times.

Answer 4

                       Map<Character,Integer> listMap = new HashMap<Character,Integer>();                         
                       Scanner in= new Scanner(System.in);
                       System.out.println("enter the string");
                       String name=in.nextLine().toString();
                       Integer value=0;
                       for(int i=0;i<name.length();i++){

                                     if(i==0){
                                                   listMap.put(name.charAt(0), 1);
                                     }
                                     else if(listMap.containsKey(name.charAt(i))){
                                                                value=listMap.get(name.charAt(i));
                                                                listMap.put(name.charAt(i), value+1);

                                     }else listMap.put(name.charAt(i),1);


                       }

                       System.out.println(listMap);

Answer 5

import java.util.Scanner;

    class Test
    { 

        static String s2="";
        int l;
        void countDuplicateCharacters(String Str)
        {
            String S=Str.toLowerCase();
            for(int i=0;i<S.length();i++)
            {
                int k=1;
                boolean value=  repeatedCheck(S.charAt(i));
                if(value==true)
                    continue;
                for(int j=i+1;j<S.length();j++)
                {


                    if(S.charAt(i)==S.charAt(j))
                    {   k++;

                    }

            }
                System.out.println("character  '" +S.charAt(i)+"' : "+k);

                s2=s2+S.charAt(i);
            }

        }

        boolean repeatedCheck(char ch)
        {
            l=s2.length();
            for (int i=0;i<l;i++)
            {
                if(s2.charAt(i)==ch)
                {
                    return true;
                }
            }

            return false;
        }

    }

    public class Duplicacy {

        public static void main(String[] args) {
            Scanner sc=new Scanner(System.in);
            System.out.println("Enter any String");
            String s=sc.nextLine();
            Test t=new Test();
            t.countDuplicateCharacters(s);


        }}

Answer 6

import java.util.HashMap;
import java.util.Scanner;


public class HashMapDemo {

    public static void main(String[] args) {
        //Create HashMap object to Store Element as Key and Value 
        HashMap<Character,Integer> hm= new HashMap<Character,Integer>();
        //Enter Your String From Console
        System.out.println("Enter an String:");
        //Create Scanner Class Object From Retrive the element from console to our java application
        Scanner sc = new Scanner(System.in);
        //Store Data in an string format
        String s1=sc.nextLine();
        //find the length of an string and check that hashmap object contain the character or not by using 
        //containskey() if that map object contain element only one than count that value as one or if it contain more than one than increment value 

        for(int i=0;i<s1.length();i++){
            if(!hm.containsKey(s1.charAt(i))){
                hm.put(s1.charAt(i),(Integer)1);
            }//if
            else{
                hm.put(s1.charAt(i),hm.get(s1.charAt(i))+1);
            }//else

        }//for

        System.out.println("The Charecters are:"+hm);
    }//main
}//HashMapDemo