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) {
        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 2

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 public class B {

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

}

}

Answer 3

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 4

//sample Input
/*2
7
saska
toro
winn
toro
vanco
saska
toro
3
effffdffffd
effffdffffd
effffdffffd*/
//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 5

I solved this with 2 dimensional array. Input: aaaabbbcdefggggh Output:a4b3cdefg4h

int[][] arr = new int[10][2];
    String st = "aaaabbbcdefggggh";
    char[] stArr = st.toCharArray();
    int i = 0;
    int j = 0;
    for (int k = 0; k < stArr.length; k++) {
        if (k == 0) {
            arr[i][j] = stArr[k];
            arr[i][j + 1] = 1;
        } else {
            if (arr[i][j] == stArr[k]) {
                arr[i][j + 1] = arr[i][j + 1] + 1;
            } else {
                arr[++i][j] = stArr[k];
                arr[i][j + 1] = 1;
            }
        }
    }
    System.out.print(arr.length);
    String output = "";
    for (int m = 0; m < arr.length; m++) {
        if (arr[m][1] > 0) {
            String character = Character.toString((char) arr[m][0]);
            String cnt = arr[m][1] > 1 ? String.valueOf(arr[m][1]) : "";
            output = output + character + cnt;

        }
    }
    System.out.print(output);

Answer 6

You can also achieve it by iterating over your String and using a switch to check each individual character, adding a counter whenever it finds a match. Ah, maybe some code will make it clearer:

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);

    }

Output:

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