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

Question

How to find the number of occurrences of a character in a string?

For example: A quick brown fox jumped over a lazy dog .

Below are some examples.

'a' = 1 'o' = 4 'space' = 8 '.' = 1 

Answer 1

 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); } } 

Exit

 Enter the Statement:asdf23123sfsdf a 1 Times s 3 Times d 2 Times f 3 Times 

Answer 2

You can use the following if String s is the string you want to process.

 Map<Character,Integer> map = new HashMap<Character,Integer>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (map.containsKey(c)) { int cnt = map.get(c); map.put(c, ++cnt); } else { map.put(c, 1); } } 

Please note that it will count all characters, not just letters.

Answer 3

  void Findrepeter(){ String s="mmababctamantlslmag"; int distinct = 0 ; for (int i = 0; i < s.length(); i++) { for (int j = 0; j < s.length(); j++) { if(s.charAt(i)==s.charAt(j)) { distinct++; } } System.out.println(s.charAt(i)+"--"+distinct); String d=String.valueOf(s.charAt(i)).trim(); s=s.replaceAll(d,""); distinct = 0; } } 

Answer 4

Java 8-way:

 "The quick brown fox jumped over the lazy dog." .chars() .mapToObj(i -> (char) i) .collect(Collectors.groupingBy(Object::toString, Collectors.counting())); 

Answer 5

It would be best to create a Map to store your account. It will be Map<Character, Integer>

You need to iterate over each character of your string and check its alphabet. You can use the Character#isAlphabetic method to do this. If it's an alphabet , increase its number in Map . If the symbol is not already in the Map , then add it with a counter of 1 .

NOTE : - The Character.isAlphabetic method is new in Java 7 . If you are using an older version, you should use Character#isLetter

  String str = "asdfasdfafk asd234asda"; Map<Character, Integer> charMap = new HashMap<Character, Integer>(); char[] arr = str.toCharArray(); for (char value: arr) { if (Character.isAlphabetic(value)) { if (charMap.containsKey(value)) { charMap.put(value, charMap.get(value) + 1); } else { charMap.put(value, 1); } } } System.out.println(charMap); 

OUTPUT : -

 {f=3, d=4, s=4, a=6, k=1} 

Answer 6

This is an implementation without using any collection and with complexity of order n. Although the decision is good enough and doesn't use Collection, it doesn't seem to care about special characters.

 import java.util.Arrays; public class DuplicateCharactersInString { public static void main(String[] args) { String string = "check duplicate charcters in string"; string = string.toLowerCase(); char[] charAr = string.toCharArray(); Arrays.sort(charAr); for (int i = 1; i < charAr.length;) { int count = recursiveMethod(charAr, i, 1); if (count > 1) { System.out.println("'" + charAr[i] + "' comes " + count + " times"); i = i + count; } else i++; } } public static int recursiveMethod(char[] charAr, int i, int count) { if (ifEquals(charAr[i - 1], charAr[i])) { count = count + recursiveMethod(charAr, ++i, count); } return count; } public static boolean ifEquals(char a, char b) { return a == b; } } 

Exit:

 ' ' comes 4 times 'a' comes 2 times 'c' comes 5 times 'e' comes 3 times 'h' comes 2 times 'i' comes 3 times 'n' comes 2 times 'r' comes 3 times 's' comes 2 times 't' comes 3 times 

Answer 7

If your string contains only alphabets, you can use something 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."); } } } } 

Conclusion:

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

Answer 8

Finding duplicates in String :

Example 1. Using a HashMap

 public class a36 { public static void main(String[] args) { String a = "Gini Rani"; fix(a); }//main public static void fix(String a ){ Map<Character ,Integer> map = new HashMap<>(); for (int i = 0; i <a.length() ; i++ ) { char ch = a.charAt(i); map.put(ch , map.getOrDefault(ch,0) +1 ); }//for List<Character> list = new ArrayList<>(); Set<Map.Entry<Character ,Integer> > entrySet = map.entrySet(); for ( Map.Entry<Character ,Integer> entry : entrySet) { list.add( entry.getKey() ); System.out.printf( " %s : %d %n" , entry.getKey(), entry.getValue() ); }//for System.out.println("Duplicate elements => " + list); }//fix } 

Example 2: using Arrays.stream () in Java 8

 public class a37 { public static void main(String[] args) { String aa = "Protijayi Gini"; String[] stringarray = aa.split(""); Map<String , Long> map = Arrays.stream(stringarray) .collect(Collectors.groupingBy(c -> c , Collectors.counting())); map.forEach( (k, v) -> System.out.println(k + " : "+ v) ); } } 

Answer 9

Use google guava Multiset<String> .

 Multiset<String> wordsMultiset = HashMultiset.create(); wordsMultiset.addAll(words); for(Multiset.Entry<E> entry:wordsMultiset.entrySet()){ System.out.println(entry.getElement()+" - "+entry.getCount()); } 

Answer 10

 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 

Answer 11

 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 12

 public static void main(String[] args) { String name="AnuvratAnuvra"; char[] arr = name.toCharArray(); HashMap<Character, Integer> map = new HashMap<Character, Integer>(); for(char val:arr){ map.put(val,map.containsKey(val)?map.get(val)+1:1); } for (Entry<Character, Integer> entry : map.entrySet()) { if(entry.getValue()>1){ Character key = entry.getKey(); Object value = entry.getValue(); System.out.println(key + ":"+value); } } } 

Answer 13

class A {

 public static void getDuplicates(String S) { int count = 0; String t = ""; for (int i = 0; i < S.length() - 1; i++) { for (int j = i + 1; j < S.length(); j++) { if (S.charAt(i) == S.charAt(j) && !t.contains(S.charAt(j) + "")) { t = t + S.charAt(i); } } } System.out.println(t); } 

}

class B open class B {

 public static void main(String[] args){ A.getDuplicates("mymgsgkkabcdyy"); } 

}

Answer 14

Using the Eclipse CharAdapter and CharBag :

 CharBag bag = CharAdapter.adapt("The quick brown fox jumped over the lazy dog.").toBag(); Assert.assertEquals(1, bag.occurrencesOf('a')); Assert.assertEquals(4, bag.occurrencesOf('o')); Assert.assertEquals(8, bag.occurrencesOf(' ')); Assert.assertEquals(1, bag.occurrencesOf('.')); 

Note: I am a committer for Eclipse collections

Answer 15

You can also do it by iterating over String and using switch to check each individual character, adding a counter when it finds a match. Ah, maybe some code will become more clear:

Main application:

 public static void main(String[] args) { String test = "The quick brown fox jumped over the lazy dog."; int countA = 0, countO = 0, countSpace = 0, countDot = 0; for (int i = 0; i < test.length(); i++) { switch (test.charAt(i)) { case 'a': case 'A': countA++; break; case 'o': case 'O': countO++; break; case ' ': countSpace++; break; case '.': countDot++; break; } } System.out.printf("%s%d%n%s%d%n%s%d%n%s%d", "A: ", countA, "O: ", countO, "Space: ", countSpace, "Dot: ", countDot); } 

Conclusion:

 A: 1 O: 4 Space: 8 Dot: 1 

Answer 16

There are three ways to find duplicates.

 public class WAP_PrintDuplicates { public static void main(String[] args) { String input = "iabccdeffghhijkkkl"; findDuplicate1(input); findDuplicate2(input); findDuplicate3(input); } private static void findDuplicate3(String input) { HashMap<Character, Integer> hm = new HashMap<Character, Integer>(); for (int i = 0; i < input.length() - 1; i++) { int ch = input.charAt(i); if (hm.containsKey(input.charAt(i))) { int value = hm.get(input.charAt(i)); hm.put(input.charAt(i), value + 1); } else { hm.put(input.charAt(i), 1); } } Set<Entry<Character, Integer>> entryObj = hm.entrySet(); for (Entry<Character, Integer> entry : entryObj) { if (entry.getValue() > 1) { System.out.println("Duplicate: " + entry.getKey()); } } } private static void findDuplicate2(String input) { int i = 0; for (int j = i + 1; j < input.length(); j++, i++) { if (input.charAt(i) == input.charAt(j)) { System.out.println("Duplicate is: " + input.charAt(i)); } } } private static void findDuplicate1(String input) { // TODO Auto-generated method stub for (int i = 0; i < input.length(); i++) { for (int j = i + 1; j < input.length(); j++) { if (input.charAt(i) == input.charAt(j)) { System.out.println("Duplicate is: " + input.charAt(i)); } } } } } 

Answer 17

 public class dublicate { public static void main(String...a) { System.out.print("Enter the String"); Scanner sc=new Scanner(System.in); String st=sc.nextLine(); int [] ar=new int[256]; for(int i=0;i<st.length();i++) { ar[st.charAt(i)]=ar[st.charAt(i)]+1; } for(int i=0;i<256;i++) { char ch=(char)i; if(ar[i]>0) { if(ar[i]==1) { System.out.print(ch); } else { System.out.print(ch+""+ar[i]); } } } } } 

Answer 18

 import java.util.HashMap; import java.util.Map; import java.util.Scanner; import java.util.Set; public class DuplicateCountChar{ public static void main(String[] args) { Scanner inputString = new Scanner(System.in); String token = inputString.nextLine(); char[] ch = token.toCharArray(); Map<Character, Integer> dupCountMap = new HashMap<Character,Integer>(); for (char c : ch) { if(dupCountMap.containsKey(c)) { dupCountMap.put(c, dupCountMap.get(c)+1); }else { dupCountMap.put(c, 1); } } for (char c : ch) { System.out.println("Key = "+c+ "Value : "+dupCountMap.get(c)); } Set<Character> keys = dupCountMap.keySet(); for (Character character : keys) { System.out.println("Key = "+character+ " Value : " + dupCountMap.get(character)); } }** 

Answer 19

In java ... use for loop:

 import java.util.Scanner; /** * * @author MD SADDAM HUSSAIN */ public class Learn { public static void main(String args[]) { Scanner sc = new Scanner(System.in); String input = sc.next(); char process[] = input.toCharArray(); boolean status = false; int index = 0; for (int i = 0; i < process.length; i++) { for (int j = 0; j < process.length; j++) { if (i == j) { continue; } else { if (process[i] == process[j]) { status = true; index = i; break; } else { status = false; } } } if (status) { System.out.print("" + process[index]); } } } } 

Answer 20

  public class StringCountwithOutHashMap { public static void main(String[] args) { System.out.println("Plz Enter Your String: "); Scanner sc = new Scanner(System.in); String s1 = sc.nextLine(); int count = 0; for (int i = 0; i < s1.length(); i++) { for (int j = 0; j < s1.length(); j++) { if (s1.charAt(i) == s1.charAt(j)) { count++; } } System.out.println(s1.charAt(i) + " --> " + count); String d = String.valueOf(s1.charAt(i)).trim(); s1 = s1.replaceAll(d, ""); count = 0; }}} 

Answer 21

 public class CountH { public static void main(String[] args) { String input = "Hi how are you"; char charCount = 'h'; int count = 0; input = input.toLowerCase(); for (int i = 0; i < input.length(); i++) { if (input.charAt(i) == charCount) { count++; } } System.out.println(count); } } 

Answer 22

 public class DuplicateValue { public static void main(String[] args) { String s = "hezzz"; char []st=s.toCharArray(); int count=0; Set<Character> ch=new HashSet<>(); for(Character cg:st){ if(ch.add(cg)==false){ int occurrences = Collections.frequency(ch, cg); count+=occurrences; if(count>1){ System.out.println(cg + ": This character exist more than one time"); } else{ System.out.println(cg); } } } System.out.println(count); } } 

Answer 23

  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 24

 import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); String reverse1; String reverse2; int count = 0; while(n > 0) { String A = sc.next(); String B = sc.next(); reverse1 = new StringBuffer(A).reverse().toString(); reverse2 = new StringBuffer(B).reverse().toString(); if(!A.equals(reverse1)) { for(int i = 0; i < A.length(); i++) { for(int j = 0; j < A.length(); j++) { if(A.charAt(j) == A.charAt(i)) { count++; } } if(count % 2 != 0) { A.replace(A.charAt(i),""); count = 0; } } System.out.println(A); } n--; } } } 

Answer 25

 public class list { public static String name(Character k){ String s="the quick brown fox jumped over the lazy dog."; int count=0; String l1=""; String l=""; List<Character> list=new ArrayList<Character>(); for(int i1=0;i1<s.length();i1++){ list.add(s.charAt(i1)); } list.sort(null); for (Character character : list) { l+=character; } for (int i1=0;i1<l.length();i1++) { if((l.charAt(i1)==k)){ count+=1; l1=l.charAt(i1)+" "+Integer.toString(count); if(k==' '){ l1="Space"+" "+Integer.toString(count); } }else{ count=0; } } return l1; } public static void main(String[] args){ String g = name('.'); System.out.println(g); } } 

Answer 26

 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]); } } 

Answer 27

 //sample Input /*2 7 saska toro winn toro vanco saska toro 3 edddddd edddddd edddddd*/ //sample output /*4 1*/ import java.util.ArrayList; import java.util.Scanner; public class MyTestWhere { /** * @param args */ public static void main(String[] args) { int count, line; Scanner sn = new Scanner(System.in); count = sn.nextInt(); sn.nextLine(); for (int i = 0; i < count; i++) { line = sn.nextInt(); sn.nextLine(); // String numArr[] = new String[line]; ArrayList<String> Arr=new ArrayList<String>(); String first = sn.nextLine(); Arr.add(first);String f; for (int j = 1; j < line; j++) { f= sn.nextLine(); for(int k=0;k<Arr.size();k++){ if(f.equalsIgnoreCase(Arr.get(k)))break; else if(k== (Arr.size()-1)){Arr.add(f);} } } System.out.println(Arr.size()); } } } 

Answer 28

 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 29

 public static void findDuplicate(String letter) { for(int i=0; i<letter.length();i++) { for( int j=i+1; j<letter.length();j++) { if(letter.charAt(i)==letter.charAt(j)) { System.out.println(letter.charAt(j)); break; } } } } 

findDuplicate ("Java");

OUTPUT: A