How to Find Duplicate Characters in String [Java Coding Problems]

Hello guys, today's programming exercise is to write a program to find repeated characters in a String. For example, if given input to your program is "Java", it should print all duplicates characters, i.e. characters appear more than once in String and their count like a = 2 because of character 'a' has appeared twice in String "Java". This is also a very popular coding question on the various level of Java interviews and written tests, where you need to write code. On the difficulty level, this question is at par with the prime numbers or Fibonacci series, which are also very popular on junior level Java programming interviews and it's expected from every programmer to know how to solve them.

I personally like this exercise because it gives beginners an opportunity to familiarize themselves with the concept of Map data structure, which allows you to store mappings in the form of key and value.

Since Map and Hash table data structure is heavily used in any enterprise Java application, good knowledge of this data structure is highly desirable among any level of Java programmers.

By the way, there are a couple of variants of this problem, which you may want to look at before going for an interview.

Sometimes an interviewer will ask you to read a file and print all duplicate characters and their count, core logic will remain the same, all you need to do is demonstrate how much you know about File IO in Java, like streaming file if it's very large rather than reading the whole file in memory.

Btw, a basic knowledge of data structure and algorithms is needed and if you need to brush up then do so. If you need a resource, I highly recommend checking out Data Structures and Algorithms: Deep Dive Using Java course on Udemy. It's a hands-on course and covers all essential data structures. It's also very affordable and you can get in just $10 on Udemy flash sales which happens every now and then.




Java Program to find Repeated Characters of String [Solution]

The standard way to solve this problem is to get the character array from String, iterate through that and build a Map with character and their count. Then iterate through that Map and print characters which have appeared more than once. So you actually need two loops to do the job, the first loop to build the map and the second loop to print characters and counts.

If you look at the below example, there is only one static method called the printDuplicateCharacters(), which does both these jobs. We first got the character array from String by calling toCharArray().

Next, we are using HashMap to store characters and their count. We use the containsKey() method to check if the key, which is a character that already exists or not already exists we get the old count from HashMap by calling the get() method and store it back after incrementing it by 1.

Once we build our Map with each character and count, the next task is to loop through Map and check each entry, if the count, which is the value of Entry is greater than 1, then that character has occurred more than once. You can now print duplicate characters or do whatever you want with them.

By the way, if you are preparing for coding interviews then I highly recommend you to join Grokking the Coding Interview: Patterns for Coding Questions course on Educative. This is an interactive, text-based coding course to learn 15 essential coding patterns like sliding window, merge interval, fast and slow pointers, etc which can be used to solve 100+ coding problems, and you can get it just for $14.9 per month membership. 

How to Find Duplicate Characters in String


And, here is the complete Java program to find duplicate characters in a given String. 
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

/**
* Java Program to find duplicate characters in String.
*
*
* @author http://java67.blogspot.com
*/
public class FindDuplicateCharacters{

    public static void main(String args[]) {
        printDuplicateCharacters("Programming");
        printDuplicateCharacters("Combination");
        printDuplicateCharacters("Java");
    }

    /*
     * Find all duplicate characters in a String and print each of them.
     */
    public static void printDuplicateCharacters(String word) {
        char[] characters = word.toCharArray();

        // build HashMap with character and number of times they appear in String
        Map<Character, Integer> charMap = new HashMap<Character, Integer>();
        for (Character ch : characters) {
            if (charMap.containsKey(ch)) {
                charMap.put(ch, charMap.get(ch) + 1);
            } else {
                charMap.put(ch, 1);
            }
        }

        // Iterate through HashMap to print all duplicate characters of String
        Set<Map.Entry<Character, Integer>> entrySet = charMap.entrySet();
        System.out.printf("List of duplicate characters in String '%s' %n", word);
        for (Map.Entry<Character, Integer> entry : entrySet) {
            if (entry.getValue() > 1) {
                System.out.printf("%s : %d %n", entry.getKey(), entry.getValue());
            }
        }
    }

}

Output
List of duplicate characters in String 'Programming'
g : 2
r : 2
m : 2
List of duplicate characters in String 'Combination'
n : 2
o : 2
i : 2
List of duplicate characters in String 'Java'


That's all on how to find duplicate characters in a String. Next time this question is asked to you in a programming job interview, you can confidently write a solution and can explain them. Remember this question is also asked to write a Java program to find repeated characters of a given String, so don't get confused yourself in wording, the algorithm will remain the same.



Related Data Structure and Algorithm Interview Questions from Javarevisited Blog
  • Top 30 Array Coding Interview Questions with Answers (see here)
  • How to reverse an array in place in Java? (solution)
  • Top 15 Data Structure and Algorithm Interview Questions (see here)
  • How to find a missing number in an array? (answer)
  • How to compare two arrays in Java? (answer)
  • Top 20 String coding interview questions (see here)
  • How to remove duplicate elements from an array in Java? (solution)
  • How to find all pairs whose sum is equal to a given number in Java (solution)
  • Top 30 linked list coding interview questions (see here)
  • Top 50 Java Programs from Coding Interviews (see here)
  • 5 Free Data Structure and Algorithms Courses for Programmers (courses)
  • 10 Algorithms Books Every Programmer Should Read (books)
  • 10 Free Data Structure and Algorithm Courses for Programmers (courses)
  • 100+ Data Structure Coding Problems from Interviews (questions)

Thanks for reading this article so far. If you like this Java coding problem and my solution then please share it with your friends and colleagues. If you have any questions or doubt then please let us know and I'll try to find an answer for you. As always suggestions, comments, innovative and better answers are most welcome.

P. S. - If you are looking for some Free Algorithms courses to improve your understanding of Data Structure and Algorithms, then you should also check the Data Structure in Java free course on Udemy. It's completely free and all you need to do is create a free Udemy account to enroll in this course. 

And, now one question for you, what is your favorite String based coding exercise? reverse String, Palindrome, or this one? or anything else, do let me know in comments. 

92 comments:

  1. How do you do it without using any additional data structure.

    ReplyDelete
    Replies
    1. import java.io.*;
      public class Str
      {
      public static void main(String arg[])throws Exception
      {
      String res="";
      BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
      String a=br.readLine();
      while(a.length()>0)
      {
      int c=0;
      for(int j=0;j<a.length();j++)
      {
      if(a.charAt(0)==a.charAt(j))
      c=c+1;
      }
      res=res+a.charAt(0)+"="+c+"\n";
      String k[]=a.split(a.charAt(0)+"");
      a=new String("");
      for(int i=0;i<k.length;i++)
      a=a+k[i];
      }
      System.out.println(res);
      }
      }

      Delete
    2. Thanks its very useful to me..

      Delete
    3. Good solution, but you can see without using data structure time complexity of your solution grows from O(n) to O(n^2). This is actually good learning point, choice of good data structure improves your algorithm.

      Delete
    4. Thanks "Anonymous", very helpful stuff for searching with minimal resources.

      Delete
    5. package StringPac;

      import java.util.Scanner;

      /**
      * @author Tarique
      * How to Print duplicate characters from String?
      */
      public class duplicateChar {

      /**
      * @param args
      */
      public static void main(String[] args)
      {

      //String str = "Nitin";

      Scanner sc = new Scanner(System.in);

      String str = sc.next();

      char d = 0;
      int count = 0;
      /* for(int i=0;i<str.length();i++)
      {

      char c = str.charAt(i);
      System.out.println(c);
      }*/
      for (int i = 0; i < str.length(); i++)
      {

      for (int j = i+1; j < str.length(); j++)
      {


      if (str.charAt(i)==str.charAt(j))
      {
      if(d!=str.charAt(i))
      {
      count++;

      d = str.charAt(i);

      System.out.println("Duplicate Charaacter is "+d);
      break;
      }

      }
      }
      }
      System.out.println("Number of duplicate character is "+count);
      }

      }

      Delete
    6. can u explain wt is d? why d!=charAt()

      Delete
    7. input is : missippi
      getting wrong output

      Delete
    8. import java.util.LinkedHashMap;
      import java.util.Map;
      import java.util.Scanner;

      public class DuplicateCharactersString {
      private static Integer getCount(char[] ch, char c, int l) {
      int cnt = 0;
      for(int i=0;i map = new LinkedHashMap();

      for(int i=0;i mapElement : map.entrySet()) {
      char key = (char)mapElement.getKey();
      int val = (int)mapElement.getValue();
      if(val>1) {
      System.out.println(key+" : "+val);
      }
      else {
      continue;
      }
      }
      sc.close();
      }
      }

      Delete
  2. Why is the 'a' of Java not occuring twice?

    ReplyDelete
    Replies
    1. @Anonymous, It actually does print 'a' twice for 'Java', its just that I missed when I copied output from Eclipse. You can try running program in your machine, it will print.

      List of duplicate characters in String 'Java'
      a : 2

      Delete
  3. How to add two arrays A={3,5,2,1} and B={6,8,9,2,4} index by index into third array C={4,5,8,9,6} answer this please..

    ReplyDelete
    Replies
    1. @Vekatesh, if all array of same length, you can use a for loop to do this job. Just iterate over both array, take each element from same index, add them and store into third array.

      Delete

    2. ///please refer below code to copy one arrays into other ,many approaches is for this but i have implemented only two
      package com.java8.demo;

      import java.lang.reflect.Array;
      import java.util.ArrayList;
      import java.util.Arrays;
      import java.util.List;
      import org.apache.commons.lang.ArrayUtils;

      public class MergeArrays {

      public static void main(String[] args) {
      // TODO Auto-generated method stub
      String a[]={"a","b","c","r","g","f"};
      String b[]={"d","e"};
      //method 1
      Object[] combined = ArrayUtils.addAll(a, b);
      System.out.println("combined arrays are:"+Arrays.toString(combined));
      //method 2
      List ls=new ArrayList<>(Arrays.asList(a));
      ls.addAll(Arrays.asList(b));
      Object o[]=ls.toArray();
      System.out.println("Merged arrays Is:"+Arrays.toString(o));
      //for sorting the arrays
      Arrays.sort(o);
      System.out.println("sort the merged ayyars:"+Arrays.toString(o));
      int d=Arrays.binarySearch(o, "b");
      System.out.println("Searched Key's index is:"+Arrays.binarySearch(o, "k")+d);

      }

      }

      Delete
  4. You can also find using a single loop.

    public static boolean checkRepeatingCharactersFromMap(String s1) {
    boolean repeat = false;
    if (s1 != null && s1.length() > 0) {
    char[] s1Array = s1.toCharArray();

    Set set = new TreeSet();
    Set repeatChar = new TreeSet();

    for (char c1: s1Array) {
    if (!set.add(c1)) {
    // System.out.print(c1 + " "); //if you want to print each occurance of the repeating character
    repeatChar.add(c1);
    repeat = true;
    // return true; //end the loop if you don't want to cache the repeating characters
    }
    }

    System.out.print(repeatChar);
    }

    return repeat;
    }

    ReplyDelete
    Replies
    1. @Genius, it will work but it will not print duplicates in the order they appear in original String. Nevertheless, if that's not stated, you can use it. Replacing TreeMap with LinkedHashMap will print the repeated character in the same order they appear in given input.

      Delete
  5. String s = "javaqjjcxcdf";
    char[] charArray = s.toCharArray();

    Map map = new HashMap();
    /*for (Character character : charArray) {
    if (map.containsKey(character)) {
    map.put(character, map.get(character) + 1);
    } else {
    map.put(character, 1);
    }
    }*/

    for (Character character : charArray) {
    map.put(character, map.get(character) != null?map.get(character)+1:1);
    }
    System.out.println(map);

    ReplyDelete
    Replies
    1. @Vikas, good choice of using ternary operator, much cleaner.

      Delete
    2. Just one changed need as the type of map should like Map map = new HashMap();

      All good, good to see example with one loop itself.

      Delete
  6. Python code:

    duplicate=[]
    first_time=[]

    a = ""Hello world"

    for i in a:
    if i not in first_time:
    first_time.append(i)
    else:
    duplicate.append(i)

    print "".join(duplicate)

    ReplyDelete
    Replies
    1. @Anonymous, Thanks for providing Python solution, neat.

      Delete
  7. l1=list('Java')
    l2=set(l1)
    [l1.remove(x) for x in l2]
    print l1

    ReplyDelete
    Replies
    1. Hello @Anonymous, Nice solution, is it Groovy, Scala or Ruby?

      Delete
  8. @Javin: How about my solution?
    My solution even prints in alphabetical order. Very much memory efficient. If the question is only to find those characters which occur more than once, then my solution runs perfectly fine.

    import java.util.*;
    import java.io.*;

    class Ideone
    {
    public static void main (String[] args) throws IOException
    {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    String s = br.readLine();
    int len = s.length();

    int[] charCount = new int[26];

    for(int i = 0; i=65 && c<=90)
    c=(char)(c+32);
    if(c>=97 && c<=122)
    charCount[c-97]+=1;
    }

    for(int i = 0; i<26; i++){
    if(charCount[i]>1)
    System.out.println((char)(i+97) + " : " + charCount[i]);
    }
    }
    }

    ReplyDelete
    Replies
    1. what is variable c ? atleast put the code without compilation errors

      Delete
    2. errors in your code

      Delete
  9. package Strings;

    import java.util.ArrayList;
    import java.util.List;

    public class PrintDuplicateStrings {
    public static void main(String[] args) {
    String s = "hello world";
    char[] chars = s.toCharArray();
    int i,j = 0;
    List finalList = new ArrayList();

    for(i=0;i<chars.length;i++){
    for(j=i+1;j<chars.length;j++){
    if(!finalList.contains(String.valueOf(chars[i])) && chars[i]==chars[j]){
    finalList.add(String.valueOf(chars[i]));
    }
    }
    }

    System.out.println("-- "+finalList);
    }
    }

    ReplyDelete
    Replies
    1. You can avoid inner loop by adding a one more condition before inner loop, that avoids inner loop if a repeated character occurs.

      public static void printDuplicateCharsInString(String str) {
      if(str==null && str.length() <=0 ) {
      return;
      }
      int length= str.length();
      List list= new ArrayList<>();
      char ch;
      int counter;
      for(int i=0; i<length ;i++ ) {
      ch=str.toLowerCase().charAt(i);
      counter=1;
      if( !list.contains(ch) && ch !=' ') {
      for(int j=i+1;j<length; j++) {
      if(ch== str.charAt(j)) {
      list.add(ch);
      counter++;
      //break;
      }
      }
      System.out.println("Character: '"+ch +"' occurs "+counter+" times");
      }
      }
      System.out.println("Duplicate Chars: "+list.toString());
      }

      Delete
  10. //Program to print duplicate characters in string
    /* package whatever; // don't place package name! */

    import java.util.*;
    import java.lang.*;
    import java.io.*;

    /* Name of the class has to be "Main" only if the class is public. */
    class Ideone
    {
    static boolean find(char[] update,char ch){
    for(int i=0;i<update.length;i++)
    if(update[i]==ch)
    return false;
    return true;
    }
    static void solution(String str){
    char[] ar=str.toCharArray();
    char[] update=new char[ar.length];
    int i=0,z=0;
    while(i<ar.length-1){
    int j=i+1;
    while(j<ar.length){
    if(find(update,ar[j])&&ar[i]==ar[j]){
    update[z++]=ar[i];
    System.out.print(ar[i]+" ");
    break;
    }
    j++;
    }
    i++;
    }
    }
    public static void main (String[] args) throws java.lang.Exception
    {
    // your code goes here
    Scanner scan=new Scanner(System.in);
    String str=scan.nextLine();
    solution(str);
    }
    }

    ReplyDelete
  11. i am getting null pointer exception how to solve it

    ReplyDelete
  12. I need to know how you print a character occurrence count Eg: Strins s = "1111223331123444" so o/p should be 14223312213143

    ReplyDelete
  13. package Arrys;

    import java.util.ArrayList;
    import java.util.Collection;
    import java.util.Collections;
    import java.util.HashMap;
    import java.util.Hashtable;
    import java.util.List;

    public class TwoStringPrograme {

    public static void main(String[] args) {
    String a="aaaxbbcccbaaasss";
    char ch;
    List cha=new ArrayList();
    int count1;
    char a1[]=a.toCharArray();
    for(int i=0;i<a1.length;i++)
    {
    count1=0;
    ch=a.charAt(i);

    if(!cha.contains(ch))
    {
    for(int j=0;j<a1.length;j++)
    {

    if(ch==a1[j])
    {
    count1++;

    }


    }
    System.out.println(ch+"--"+count1);
    cha.add(ch);
    }
    }

    } }

    ReplyDelete
  14. // without using any buildin function

    public static void main(String[] args) {
    char[] array = "aabsbdcbdgratsbdbcfdgs".toCharArray();
    char[][] countArr = new char[array.length][2];
    int lastIndex = 0;
    for (char c : array) {
    int foundIndex = -1;
    for (int i = 0; i < lastIndex; i++) {
    if (countArr[i][0] == c) {
    foundIndex = i;
    break;
    }
    }
    if (foundIndex >= 0) {
    int a = countArr[foundIndex][1];
    countArr[foundIndex][1] = (char) ++a;
    } else {
    countArr[lastIndex][0] = c;
    countArr[lastIndex][1] = '1';
    lastIndex++;
    }
    }
    for (int i = 0; i < lastIndex; i++) {
    System.out.println(countArr[i][0] + " " + countArr[i][1]);
    }
    }

    ReplyDelete
  15. public void printDups(String str) {

    if(str != null && str.length() > 0) {
    char [] dupArr = str.toCharArray();

    System.out.println("Dup Array ");
    String dups ="";
    for(int i=0; ii; j--){
    if (dupArr[i] == dupArr[j]) {
    dups = (dups+dupArr[i]).trim();
    System.out.print(dupArr[i]);
    }
    }
    }

    }
    }

    ReplyDelete
  16. Instead of clumsy solutions above, just loop the string once from left to right so to speak and at every character check if it also exists on the right hand side yet left to loop. A one liner and much easier to read. Work smarter, not harder!

    ReplyDelete
  17. #python Code
    import sys

    myStr='java'
    #try with javaav

    first_time=[]
    duplicate=[]

    #Empty String
    if len(myStr)<1:
    print "empty string"
    sys.exit()

    #iterating over the string
    for i in myStr:
    if i not in first_time:
    first_time.append(i)
    else:
    if i not in duplicate:
    duplicate.append(i)
    print "".join(duplicate)

    output: a
    output 2nd test case: av
    Credits- above anonymous coder

    ReplyDelete
  18. class Practise3

    {

    public static void main(String[] args) throws ArrayIndexOutOfBoundsException
    {

    for(int i=0;i<args.length;i++)
    {
    System.out.println("Enter the string :" + args[i]);
    String s = (String)args[i];
    System.out.println(s);
    try
    {
    for( int k =0; k<s.length();k++)
    {
    for(int j = k+1; j!=s.length();j++)
    {
    char var = s.charAt(k);
    if(var == s.charAt(j))
    {
    System.out.println("duplicate character is : " + s.charAt(j));
    }
    else
    {
    System.out.println(" no duplicates found" + ".." + s.charAt(k));
    }
    }
    }
    }
    catch (Exception e)
    {
    System.out.println("no");
    }




    }

    }
    }
    I just tried using String class and its methods

    ReplyDelete
  19. Why not recursion? It's powerful and logical.

    public class StringDuplicate {
    //search from the beginning of the string
    public static void printDupes(String string) {
    //specify exit case, when we finish the whole string
    if(string.length() == 0) return;
    //find the string in which the character you're searching for is dropped
    String substring = string.substring(1);
    //if we found that character on the substring, we sysout that b**ch
    if(substring.indexOf(string.charAt(0)) > 0) {
    System.out.println(string.charAt(0));
    }
    //we let the recursion go to the shorter string...
    //by the process of induction, this will always terminate ;)
    printDupes(substring);
    }
    public static void main(String[] args) {
    printDupes(args[0]);
    }
    }

    ReplyDelete
  20. Why not recursion? Less storage, sleek, unique, and cool! :)
    public class StringDuplicate {
    public static void printDupes(String string) {
    //exit condition
    if(string.length() == 0) return;
    //find the substring in which to see if the first character is in
    String substring = string.substring(1);
    //what you do if you find it? print it!
    if(substring.indexOf(string.charAt(0)) > 0) {
    System.out.println(string.charAt(0));
    }
    //recusion on that b*tch
    printDupes(substring);
    }
    public static void main(String args[]) {
    printDupes(args[0]); //utilize command line arguments for general cases
    }
    }

    ReplyDelete
    Replies
    1. @artsArt, yes, you can use recursion, no problem with that, but your solution will only print the duplicate characters, not their count. How can you improve the solution to print the count of duplicate character?

      Delete
  21. checks for words separated by space, removers symbols too.

    def removechar(dup):
    symbol = "~`!@#$%^&*()_-+={}[]:>;',</?*-+"
    for i in dup:
    if i in symbol or i is ' ':
    dup.remove(i)
    return dup
    def duplicate(str):
    l = []
    dup = []
    str = list(str)
    for i in str:
    try:
    j = list(i)
    for k in j:
    if k not in l:
    l.append(k)
    else:
    dup.append(k)
    except:
    print "No Duplicate"
    if len(dup) == '0':
    print "No Dupliates"
    else:
    d = []
    final = dup
    for i in dup:
    if i not in d:
    d.append(i)
    print "Total numbe of duplicates:",len(removechar(d)),"\nThey are:",','.join(removechar(d))
    if __name__ == '__main__':
    duplicate(raw_input("Enter the string"))

    ReplyDelete
  22. public static void printDuplicatechar(String s) {

    char[] arr = s.toCharArray();

    List list=new ArrayList<>();


    for (int i = 0; i < arr.length; i++) {
    int h = 0;
    for (int j = 0; j < arr.length; j++) {
    if (arr[i] == arr[j]) {

    h++;
    if (h >= 2) {
    list.add(arr[i]);

    }
    }
    }
    }
    list.stream().distinct().forEach(e->System.out.println(e));
    }

    ReplyDelete
  23. Map map = Arrays.stream(s.split("")).collect(Collectors.groupingBy(c -> c , Collectors.counting()));
    List duplicates = map.entrySet().stream().filter(entry -> entry.getValue() > 1).map(k -> k.getKey()).collect(Collectors.toList());

    ReplyDelete
  24. package logic;
    public class DuplicateChar {
    public static void findDuplicateCharacters(String input1)
    {
    int i;

    char[] a=input1.toCharArray();
    int h[]=new int[26];
    for(i=0;i64&&a[i]<95)
    {
    h[a[i]-65]++;
    }
    if(a[i]>95&&a[i]<123)
    {
    h[a[i]-97]++;
    }
    }
    for(i=0;i64&&a[i]<95)
    {
    if(h[a[i]-65]>1)
    System.out.println(a[i]+" "+h[a[i]-65]);
    h[a[i]-65]=0;
    }

    if(a[i]>95&&a[i]<123)
    {
    if(h[a[i]-97]>1)
    System.out.println(a[i]+" "+h[a[i]-97]);
    h[a[i]-97]=0;
    }
    }
    }
    public static void main(String args[])
    {

    findDuplicateCharacters("programming");
    }
    }

    ReplyDelete
  25. public class Duplicatechar {
    public static void main(String[] args) {
    String s ="java";
    char x;
    int i;
    int c=0;
    String str2=s.substring(1);

    for(i=0;i<s.length();i++)
    {
    c=0;
    x=s.charAt(i);
    for(int j=0;j<str2.length();j++)
    {
    char y=str2.charAt(j);
    if(x==y)
    {
    c++;

    if(c==2)
    System.out.println("the duplicate char "+x);
    }
    }

    }

    }

    }
    I have one mistake it print for me twice the comment that the duplicate character is a .. can anyone help me .. thanks

    ReplyDelete
  26. public class RemoveDup
    {
    public static void main(String args[])
    {
    String s;
    int i;
    Scanner sc=new Scanner(System.in);
    System.out.println("Enter any String");
    s=sc.next();
    char[] chars= s.toCharArray();
    Set charSet = new LinkedHashSet();
    for (char c : chars)
    {
    charSet.add(c);
    }
    StringBuilder sb = new StringBuilder();//to buffer characters
    for (Character character : charSet)
    {
    sb.append(character);
    }
    System.out.println(sb.toString());
    }
    }

    ReplyDelete
  27. #include
    #include

    void main()
    {

    char s[100],ch;
    int i,j,n,flag=0;

    printf("Enter\n");
    scanf_s("%s",s,8);

    n=strlen(s);

    printf("%d\n", n);

    for(i=0;i=2)
    {

    printf("%c\n", ch);

    }
    else
    {
    printf("No\n");

    }



    }


    ReplyDelete
  28. #include
    #include
    #include

    using namespace std;

    struct node
    {
    char ch;
    int count;
    struct node *next;
    };
    void main ()
    {
    node * Head = NULL;
    node * crawl = Head;
    char input[100];
    memset(input,0,100);
    printf("Enter the string :- \n");
    scanf("%[^\n]s",&input);

    if(input == NULL)
    {
    printf("NULL String!!!");
    getch();
    return;
    }
    int nLen = strlen(input);
    if(nLen <= 1)
    {
    printf("Invalid String!!!");
    getch();
    return;
    }

    for(int n = 0; n 0)
    {
    if(Head == NULL)
    {
    Head = new node();
    Head->next = NULL;
    Head->ch = ref;
    Head->count = nCount;
    }
    else
    {
    crawl = Head;
    while (crawl->next != NULL && crawl->ch != ref)
    {
    crawl= crawl->next;
    }

    if(crawl->next == NULL)
    {
    if(crawl->ch != ref)
    {
    crawl->next = new node();
    crawl = crawl->next;
    crawl->next = NULL;
    crawl->ch = ref;
    crawl->count = nCount;
    }
    }
    }
    }

    }

    while(Head != NULL)
    {
    printf("%c is duplicated %d times\n",Head->ch,Head->count);
    crawl = Head;
    Head = Head->next;

    delete crawl;
    }
    }

    ReplyDelete
  29. #include
    #include
    #include

    using namespace std;

    struct node
    {
    char ch;
    int count;
    struct node *next;
    };
    void main ()
    {
    node * Head = NULL;
    node * crawl = Head;
    char input[100];
    memset(input,0,100);
    printf("Enter the string :- \n");
    scanf("%[^\n]s",&input);

    if(input == NULL)
    {
    printf("NULL String!!!");
    getch();
    return;
    }
    int nLen = strlen(input);
    if(nLen <= 1)
    {
    printf("Invalid String!!!");
    getch();
    return;
    }

    for(int n = 0; n 0)
    {
    if(Head == NULL)
    {
    Head = new node();
    Head->next = NULL;
    Head->ch = ref;
    Head->count = nCount;
    }
    else
    {
    crawl = Head;
    while (crawl->next != NULL && crawl->ch != ref)
    {
    crawl= crawl->next;
    }

    if(crawl->next == NULL)
    {
    if(crawl->ch != ref)
    {
    crawl->next = new node();
    crawl = crawl->next;
    crawl->next = NULL;
    crawl->ch = ref;
    crawl->count = nCount;
    }
    }
    }
    }

    }

    while(Head != NULL)
    {
    printf("%c is duplicated %d times\n",Head->ch,Head->count);
    crawl = Head;
    Head = Head->next;

    delete crawl;
    }
    }

    ReplyDelete
  30. See following solution with O(n) complexity.

    public static void FindDuplicateCharacters(String str){

    if(str == null || str.length() == 0){
    System.out.println("Invalid String");
    return;
    }

    int[] counter = new int[128];
    for(char c : str.toCharArray()){
    counter[c]++;
    }

    for(int i = 0 ; i < str.length(); i++){
    char c = str.charAt(i);
    if(counter[c] > 1){
    System.out.println(c);
    counter[c] = 1;
    }
    }
    }

    ReplyDelete
  31. a=raw_input("Enter The String")

    print set([x for x in a if a.count(x)>1])


    In Python

    ReplyDelete
  32. System.out.println("Please Enter Your String");
    Scanner sc = new Scanner(System.in);
    String inputString = sc.next();
    System.out.println("Your String ::: " + inputString);

    if(null != inputString && inputString.length() > 0) {
    for(int i = 0; i < inputString.length(); i++) {
    int j = i;
    while(j > 0) {
    if(inputString.charAt(i) == inputString.charAt(j-1) )
    System.out.println(inputString.charAt(i));
    j--;
    }
    }
    }
    sc.close();

    ReplyDelete
  33. I will use map.put(c[i], map.getOrDefault(c[i], 1) + 1);
    to avoid else condition

    ReplyDelete
    Replies
    1. Hello Nitesh, that is good but how about getOrDefault() method, do you know from which Java version it is available?

      Delete
  34. i'm beginner so my solution maybe is not rational. I tried it in c++

    #include
    #include
    using namespace std;

    void main()
    {
    char s[20], *p, s_[2]="", s1[20]="";
    cin >> s;
    p=s;
    while (strlen(p)>0)
    {
    strncpy_s(s_, p, 1);
    strncpy_s(s1, s, strlen(s)-strlen(p));
    if(strrchr(p, *p)!=strchr(p, *p)
    && strchr((p+strcspn(p, s_)+1), *p)==strrchr(p, *p)
    && strcspn(s1, s_)==strlen(s1) )
    cout << *p << ": " << 2 << endl;
    p++;
    }
    system("pause");
    }

    ReplyDelete
  35. import java.util.Scanner;

    public class DuplicateString {

    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    char[] str = sc.next().toLowerCase().toCharArray();

    int[] charArray = new int[26];

    for (int i = 0; i < str.length; i++) {
    charArray[Math.abs(97-str[i])]++;
    }

    for (int i = 0; i < charArray.length; i++) {
    if (charArray[i] > 1) {
    System.out.printf("%c",Math.abs(97+i));
    System.out.println();
    }
    }
    }
    }

    ReplyDelete
  36. #include

    2

    #include

    3

    #include

    4

    #include

    5



    6

    int main() {

    7

    char a[10];

    8

    int i,j,has[1000]={0},l;

    9

    scanf("%s",a);

    10

    l=strlen(a);

    11

    for(i=0;i1)

    17

    {

    18

    printf("%c -> %d\n",i+65,has[i]);

    19

    }

    20

    }

    21



    22

    }

    23

    ReplyDelete
  37. private static void characterCount(String str) {
    HashMap charCountMap = new HashMap<>();
    for (Character ch : str.toLowerCase().toString().toCharArray()) {
    Integer count = charCountMap.putIfAbsent(ch, 1);
    if (null != count) {
    charCountMap.put(ch, charCountMap.get(ch) + 1);
    }
    }
    Map charMap =
    charCountMap.entrySet().stream().filter(map -> map.getValue() != 1).collect(Collectors.toMap(p -> p.getKey(), p -> p.getValue()));
    System.out.println(charMap);
    }

    ReplyDelete
  38. public class RemoveDuplicatesFromString {

    public static void main(String[] args) {

    String input = "hello world";
    String output = removeDuplicates(input);
    System.out.println(output);
    }

    private static String removeDuplicates(String input) {
    HashSet mySet = new HashSet();
    String output="";

    for(Character c :input.trim().replaceAll(" ", "").toCharArray())
    {
    if(mySet.add(c)==false)
    if(!output.contains(c+""))
    output = output+c;
    }
    return output;
    }
    }

    ReplyDelete

  39. import java.util.Scanner;

    public class JavaApplication19 {
    public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    System.out.println("Program to print repeated characters in a string");
    System.out.println("Enter the string");
    String s1=sc.next();
    System.out.println("Repeated characters in the given string is:");
    for(int i=0;i=2){
    System.out.print(s1.charAt(i));
    }
    }
    }
    }
    System.out.println();
    }
    }

    ReplyDelete
  40. import java.util.Scanner;
    public class JavaApplication19 {
    public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    System.out.println("Program to print repeated characters in a string");
    System.out.println("Enter the string");
    String s1=sc.next();
    String s2="";
    int flag=0;
    for(int i=0;i=2){
    flag=1;
    s2=s2+String.valueOf(s1.charAt(i));
    }
    }
    }
    }
    if(flag==1){
    System.out.println("Repeated characters in the given string is:");
    System.out.println(s2);
    }else{
    System.out.print("NULL");
    }
    System.out.println();
    }
    }

    ReplyDelete
  41. import java.util.List;
    import java.util.ArrayList;
    import java.util.Scanner;
    public class S5 {
    static int k=0,val=0;
    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String s1=sc.next();
    int count=0;
    String otherString="";
    List l1= new ArrayList();
    for( char i='a';i<='z';i++){

    count = s1.length()-s1.replaceAll( String.valueOf(i) ,"").length();
    // System.out.println(count);
    if(count>1){
    String character = ""+i;
    otherString = otherString.concat(character);

    l1.add(count);

    }
    }
    char[] ch=otherString.toCharArray();
    int[] ret = new int[l1.size()];
    int j = 0;
    for (Integer e : l1)
    ret[j++] = e.intValue();
    for(int i=0;i<ch.length;i++)
    {
    System.out.println(ch[i]+": "+ret[i]);
    }

    }
    }

    ReplyDelete
  42. public class Duplicatechar{
    //How do you do it without using any additional data structure.
    public static void main(String[] args)
    {
    String name= "javava", rev="";
    for(int i=0; i<name.length(); i++)
    {
    int count =1;
    for(int j=0; j<name.length(); j++){
    if(i != j){
    while(name.charAt(i)==name.charAt(j)){
    count++;
    break;
    }
    }
    }
    if(rev.indexOf(name.charAt(i))!=-1){
    System.out.print("");
    }else{
    rev += String.valueOf(name.charAt(i));
    System.out.println("character is ="+name.charAt(i)+" and Count is = "+count);
    }
    }

    }
    }

    ReplyDelete
  43. U can add counter to count if 2 or more duplicates are in that word and store it in list as well. List is optional just for easy lookup for duplicates.
    Simple way:-->

    import java.util.ArrayList;
    import java.util.List;

    public class Multy {

    public static void main(String[] args){
    String[] words = {"Programming","Karma","Colleague"};
    FindMultiplications(words[0]);
    }
    static void FindMultiplications(String words){
    char[] h = words.toCharArray();
    List cont = new ArrayList<>();
    for (int i=0;i<words.length();i++){
    for (int j=0;j<words.length();j++){
    if (i==j) continue;
    else if (h[i]==h[j] && !(cont.contains(h[i]))){
    System.out.println("Duplicate exist "+h[i]+" "+h[j]);
    cont.add(h[i]);
    }
    else System.out.println("Print "+h[i]+" "+h[j]);
    }
    }
    }
    }

    ReplyDelete
  44. package com.stringprac;

    import java.util.Scanner;

    public class CharCount {

    public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    System.out.println("please enter a string");
    String s = sc.nextLine();
    int count=1;
    for(int i=0; i<s.length(); i++) {
    for(int j=i+1; j<s.length(); j++) {
    if(s.charAt(i)==s.charAt(j)) {
    count++;
    System.out.println(s.charAt(i)+": "+count);
    }
    }
    count=1;
    }
    }
    }

    ReplyDelete
  45. If you want only to print duplicate characters...
    Here, I'm iterating through only half of the String. Suggestions for improvement are always welcome :)

    // Print duplicate characters of a String.
    public static void printDuplicates(String str) {
    if (str.length() == 1) {
    System.out.println("No duplicate found!");
    return;
    }
    char[] arr = str.toCharArray();
    Set setOfDuplicates=new HashSet();
    int length = arr.length;
    List list=new ArrayList();
    for (int i = 0; i < (length) / 2; i++) {
    System.out.println("i= "+i);

    if(list.contains(arr[i]))
    setOfDuplicates.add(arr[i]);
    else
    list.add(arr[i]);
    if(list.contains(arr[length-i-1]))
    setOfDuplicates.add(arr[length-i-1]);
    else
    list.add(arr[length-i-1]);
    }
    System.out.println(setOfDuplicates);
    }

    ReplyDelete
  46. //Print repetive characters from string.
    public static void GetDuplicateChars()
    {
    string str="TestMyTesty";
    for(int i=0;i-1)
    Console.Write(str[i]);
    }
    }

    ReplyDelete
    Replies
    1. public static void main(String[] args) {
      String input = "kdsfglčjretg945ghdčfsgfjki235";
      Map M = new HashMap<>();
      for (char c : input.toCharArray()) {
      String s = String.valueOf(c);
      M.compute(s, (String key, Integer value) -> {
      return value == null ? 1 : 1 + value;
      });
      }

      System.err.println(M);
      }

      Delete
  47. Given a string print only repeated individual
    characters of the string in the order they occur

    if no such characters occur print -1

    input =
    abcac
    output =
    acac


    input =
    ahdaa
    output =
    aaa

    input =
    ijkl
    output =
    -1

    ReplyDelete
  48. public Set getDuplicates(String string) {
    Set toReturn = new LinkedHashSet<>();
    if (string == null || string.isEmpty()) {
    return toReturn;
    } else {
    for (int i = 0; i < string.length(); i++) {
    Character c = string.charAt(i);
    if (string.substring(i + 1, string.length()).contains(c.toString())) {
    toReturn.add(c);
    }
    }

    return toReturn;
    }

    }

    ReplyDelete
  49. This is my version of this problem
    public static char duplicateCharInString(String input) {
    char ch =' ';
    Set nonDuplicate = new HashSet();
    for(int i=0;i<input.length();i++) {
    if(!nonDuplicate.contains(input.charAt(i))){
    nonDuplicate.add(input.charAt(i));
    }
    else {
    return input.charAt(i);
    }
    }
    return ch;
    }

    ReplyDelete
  50. C/C++ Solution:

    #include
    #include

    using namespace std;

    void getDuplicate(char ch[]);
    int main()
    {
    cout << "\n String Test Program";
    char ch[20];
    cout << "\nEnter input string: ";
    cin >> ch;

    getDuplicate(ch);

    _getch();
    return 0;
    }

    void getDuplicate(char ch[])
    {
    int Arr[255] = { 0 };
    for (int i = 0; i < strlen(ch); i++)
    {
    Arr[ch[i]] += 1;
    }

    for (int i = 0; i < 255; i++)
    {
    if (Arr[i] >= 2)
    cout << char(i) << " ";
    }
    }

    ReplyDelete
  51. `` public static String printDuplicate(String s) {
    int counts[] = new int[256];
    StringBuilder res = new StringBuilder();

    for (char c : s.toCharArray()) counts[c]++;

    for (int i = 0; i < counts.length; i++) {
    if (counts[i] > 1) res.append((char) i);
    }
    return res.toString();
    }
    ``

    ReplyDelete
    Replies
    1. Good job. did you tested it for null and empty values? though, not required here, you should always handle edge cases, now a days many companies using interview portals to check your solution which has many test cases with boundary conditions. Though this solution is all right it will not pass those edge cases.

      Delete
  52. Anybody knows how to find and remove duplicate words in as string with concept of hashtable and string tokenizer

    ReplyDelete
  53. My version of code

    #include
    using namespace std;
    int main()
    {
    string str;
    getline(cin,str);

    for(int i=0;i<str.length()-1;i++){
    for(int j=i+1;j<str.length();){
    if(str.at(i)==str.at(j)){
    cout<<"The Repeated Character is "<<str.at(j)<<endl;
    break;
    }
    else
    {j++;
    }
    }

    }
    }

    ReplyDelete

  54. public class CountChar2 {
    public static void main(String[] args) {

    String str= "javasixtyseven";

    char[] ch=str.toCharArray();
    int len=ch.length;

    for (int i = 0; i < ch.length; i++) {
    int count =0;
    char c=str.charAt(i);

    for(int j = 0;j0){
    System.out.println(c+" "+count);
    }
    }


    }// main
    }// class

    ReplyDelete
  55. #include
    #include
    int main()
    {
    char s[100];
    int cnt[10001];
    int i,j,k;
    printf("enter string\n");
    gets(s);
    k=strlen(s);
    for(i=0;i='a'&&s[i]<='z')
    cnt[s[i]-'a']++;
    else if(s[i]>='A'&&s[i]<='Z')
    cnt[s[i]-'A']++;
    }

    for(i=0;i<26;i++)
    {
    if(cnt[i]>1)
    printf("%c :%d\n",i+'a',cnt[i]);
    }

    return 0;
    }

    ReplyDelete
  56. S="String with duplicates";
    p=S.charAt(0);
    for(i=1;i<S.length;i++){
    c=p|S.charAt(i);
    if(p==c)
    { do whatever you want to do}
    p=c;
    }

    ReplyDelete
  57. C method:

    #include
    #include
    const char* str = NULL;
    int main(int argc, const char** argv){
    str = argv[1];
    printf("Original String: %s\n", str);
    for(int i = 0; i < strlen(str); i++){
    for(int j = i+1; str[j] != '\0'; j++){
    if(str[i] == str[j]){
    printf("duplicate: %c\n", str[i]);
    }
    }
    }
    }

    ReplyDelete
  58. import java.util.HashMap;
    import java.util.Map;

    public class PrintDuplicateCharacters2 {

    public static void main(String[] args) {


    String s="Hello check the duplicate strings";

    Map mp=new HashMap();

    char c[]=s.toCharArray();

    for(int i=0;i1){
    System.out.println(c[i]+" is duplicate "+((int)(mp.get(c[i]) )-1)+" times");
    }
    }
    }
    }

    ReplyDelete
  59. import java.util.*;

    public class StringDuplicate {

    public static void main(String[] args) {
    // TODO Auto-generated method stub
    String s = "hellohello";
    char[] charArray = s.toCharArray();

    Map map = new HashMap();

    for (Character character : charArray) {
    map.put(character, map.get(character) != null? map.get(character)+1:1);
    }
    System.out.println(map);

    }

    }

    ReplyDelete
  60. import java.util.*;

    public class StringDuplicate {

    public static void main(String[] args) {
    // TODO Auto-generated method stub
    String s = "hellohello";
    char[] charArray = s.toCharArray();

    Map map = new HashMap();

    for (Character character : charArray) {
    map.put(character, map.get(character) != null? map.get(character)+1:1);
    }
    System.out.println(map);

    }

    }

    ReplyDelete
  61. public static voi main(String[] args){


    String name= "sachast";
    char temp = 0;
    int cnt=0;
    char[] duplicatechars=name.toCharArray();

    for (int i = 0; i < duplicatechars.length; i++) {
    Boolean flag=false;

    for (int j = i+1; j < duplicatechars.length; j++) {
    if(duplicatechars[i]==duplicatechars[j]) {
    flag=true;
    cnt++;
    temp=duplicatechars[i];
    }
    }

    if(flag) {
    System.out.println("Charcter:"+temp +"\n"+"Number of times it occures:"+cnt);
    }
    cnt--;
    }
    }

    ReplyDelete
  62. public static void main(String[] args) {
    String string = "Programming";
    int counter = 0;
    Map characters = new HashMap();
    char chars[] = string.toUpperCase().toCharArray();
    for (int i = 0; i < chars.length; i++) {

    for (int j = 0; j < chars.length; j++) {
    if (chars[i] == chars[j]) {
    counter++;

    }
    }
    characters.put(chars[i],counter);
    counter = 0;
    }
    //Java 8 Streams
    // It will collect all the values whose letters are repeated more than once
    characters.entrySet().stream().filter(x->x.getValue()>1).collect(Collectors.toList())
    //Printing all the letters which are repated more than once
    .stream().forEach(System.out::println);
    }

    ReplyDelete
  63. function dubStr(str) {
    var copy = "", result = {};
    for(var i = 0; i <= str.length - 1; i++) {
    if(copy.indexOf(str[i]) === -1) {
    copy = copy.concat(str[i]);
    } else {
    var counter = (result[str[i]] !== undefined) ? result[str[i]]+1 : 2;
    result[str[i]] = counter;
    }
    }
    return result;
    }

    dubcateStr("Javavkokkkvvaaa") // {a: 5, v: 4, k: 4}

    ReplyDelete
  64. Before converting the String to character array , we should convert it to the Lowercase/Uppercase format.

    ReplyDelete
  65. this will give null pinter exception

    ReplyDelete
  66. package com.telekom.de;

    import java.util.Arrays;

    public class PrintDuplicates {

    public static void main(String[] args) {

    String str = "Programming";
    str = str.toLowerCase();
    char[] charArray = str.toCharArray();

    Arrays.sort(charArray);

    char previous = charArray[0];
    int counter = 1;
    for(int i = 1; i < charArray.length; i++){
    if(previous == charArray[i]){
    counter++;
    if(counter > 1) System.out.printf("previous : %s, coun : %d", previous, counter);
    } else {
    counter = 1;
    previous = charArray[i];
    }
    }

    }

    }

    ReplyDelete

Feel free to comment, ask questions if you have any doubt.