CODEWITHSUNDEEP
java
Ankesh Pandey
Ankesh Pandey

Reputation: 157

Find the occurrence of each character in a given string

"I want to find and print the occurrence of each character of given string and i have build my own logic but there is some problem.for example if i gave input as 'JAVA'. the output that my program produce will be 

J 1
A 2
V 1
A 1

Expected output :

    J 1
    A 2
    V 1

i doesn't want to print A again. I hope you all get it what is the problem in my code."

import java.util.Scanner;

public class FindOccuranceOfCharacter {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    String x;
    Scanner input = new Scanner(System.in);
    System.out.println("Enter a string");
    x = input.nextLine();
    x = x.toUpperCase();
    int size = x.length();
    for(int i =0;i<size;i++) {
        int count=1;
        char find = x.charAt(i);

        for(int j=i+1;j<size;j++) {
            if(find == x.charAt(j)) {
                count++;
            }
        }

        System.out.printf("%c\t%d",x.charAt(i),count);
        System.out.println();       
    }
}

}

Upvotes: 0

Views: 958

Answers (3)

Matthias
Matthias

Reputation: 7521

This is not an optimal solution, but I have tried to change your code as little as possible:

public static void main(String args[]) {
    Scanner input = new Scanner(System.in);
    System.out.println("Enter a string");
    // use a StringBuilder to delete chars later on
    StringBuilder x = new StringBuilder(input.nextLine().toUpperCase());
    for(int i=0;i<x.length();i++) {
        int count=1;
        char find = x.charAt(i);

        // go through the rest of the string from the end so we do not mess up with the index
        for(int j=x.length()-1;j>i;j--) {
            if(find == x.charAt(j)) {
                count++;
                // delete counted occurences of the same char
                x.deleteCharAt(j);
            }
        }

        System.out.printf("%c\t%d",x.charAt(i),count);
        System.out.println();       
    }
}

My more preferred Java stream would look like this:

input.nextLine().toUpperCase().chars()
        .mapToObj(i -> (char) i)
        .collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()))
        .forEach((k, v) -> System.out.println(k + "\t" + v));

Upvotes: 1

Fabrizio R.
Fabrizio R.

Reputation: 126

Using hashMap it's easy to accumulate the number of occurrences and you can easily print iterating the HashMap.

This is the code: 

public class FindOccuranceOfCharacter {

public static void main(String[] args) {

    String x;
    Scanner input = new Scanner(System.in);
    System.out.println("Enter a string");
    x = input.nextLine();
    HashMap<Character,Integer> occurance = new HashMap<Character,Integer>();

    x = x.toUpperCase();
    int size = x.length();
    for(int i =0;i<size;i++) {
        int count=1;
        char find = x.charAt(i);

        occurance.put(find, occurance.getOrDefault(find, 0) + 1);

    }

    for (Character key : occurance.keySet()) {
        Integer value = occurance.get(key);
        System.out.println("Key = " + key + ", Value = " + value);
    }

}

Upvotes: 1

Elliott Frisch
Elliott Frisch

Reputation: 201537

The reason your code prints the way it does is that your loop prints each character (and subsequent matches) for a given index. You really need to store the character and counts in a data structure with one loop, and then display the counts with a second. A LinkedHashMap<Character, Integer> is perfect for your use case (because it preserves key insertion order, no additional logic is needed to restore input order). Additional changes I would make include using String.toCharArray() and a for-each loop. Like,

Map<Character, Integer> map = new LinkedHashMap<>();
for (char ch : x.toUpperCase().toCharArray()) {
    map.put(ch, map.getOrDefault(ch, 0) + 1);
}
for (char ch : map.keySet()) {
    System.out.printf("%c\t%d%n", ch, map.get(ch));
}

Which I tested with x equal to JAVA and got (as requested)

J   1
A   2
V   1

Upvotes: 4

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