In this tutorial, we will see simple program to find frequency of Each Character in a String.
There are multiple ways to solve this proble. Let’s see one by one.
Table of Contents
Using counter array
Here is the algorithm for the same.
- Initialize
counterarray of 256 length - Iterate over
Stringand increase count by 1 at index based onCharacter. For example: If we encounter ‘a’ in String, it will be likecounter[97]++as ASCII value of ‘a’ is 97. - Iterate over
counterarray and print character and frequency ifcounter[i]is not 0.
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import java.util.Scanner; public class StringOperator { public static void main(String args[]) { int i; String str; int counter[] = new int[256]; Scanner in = new Scanner(System.in); System.out.print("Enter a String : "); str=in.nextLine(); for (i = 0; i < str.length(); i++) { counter[(int) str.charAt(i)]++; } // Print Frequency of characters for (i = 0; i < 256; i++) { if (counter[i] != 0) { System.out.println("The character " + (char) i + " has occurred for " + counter[i] + " times"); } } } } |
Output:
Enter a String : Java2Blog
The character B has occurred for 1 times
The character J has occurred for 1 times
The character a has occurred for 2 times
The character g has occurred for 1 times
The character l has occurred for 1 times
The character o has occurred for 1 times
The character v has occurred for 1 times
The character B has occurred for 1 times
The character J has occurred for 1 times
The character a has occurred for 2 times
The character g has occurred for 1 times
The character l has occurred for 1 times
The character o has occurred for 1 times
The character v has occurred for 1 times
Using HashMap
We can use HashMap as well to find Frequency of Each Character in a String.
- Create a HashMap which will contain character to count mapping.
- Iterate over String
- make count to 1 if
HashMapdo not contain thecharacterand put it in HashMap with key as Character and count as value - If
HashMapalready have thecharacter, increment its count by1
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package org.arpit.java2blog; import java.util.HashMap; public class FindFrequencyCharMain { public static void main(String[] args) { String str = "java2blog"; HashMap<Character,Integer> charFreqMap = new HashMap<>(); for(int i= 0 ; i< str.length() ; i++) { Character ch=str.charAt(i); if(charFreqMap.containsKey(ch)) { int count = charFreqMap.get(ch); charFreqMap.put(ch,count+1); } else { charFreqMap.put(ch,1); } } System.out.println(charFreqMap); } } |
Output:
{a=2, 2=1, b=1, v=1, g=1, j=1, l=1, o=1}
Using HashMap’s computeIfPresent and computeIfAbsent[java 8]
We can also use HashMap’s computeIfPresent and computeIfAbsent to write code in more functional way. It uses Java 8‘s lambda expressions to write the logic.
Let’s rewrite the code as below.
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package org.arpit.java2blog; import java.util.HashMap; public class FindFrequencyCharMain { public static void main(String[] args) { String str = "java2blog"; HashMap<Character,Integer> charFreqMap = new HashMap<>(); for(int i= 0 ; i< str.length() ; i++) { Character ch=str.charAt(i); charFreqMap.computeIfPresent(ch, (character,count)-> count+1); charFreqMap.computeIfAbsent(ch, (character)-> 1); } System.out.println(charFreqMap); } } |
It will generate same output as above program.
That’s all about Java Program to Find Frequency of Each Character in a String.
