Java Count Characters Not Next to Whitespace

Day 9 Java Practice: Find the First Non-Repeated Character in a String While practicing Java, I worked on a classic string problem: 👉 Find the first non-repeated character in a given string. For example, in the string "swiss", the first character that does not repeat is 'w'. To solve this, I used a LinkedHashMap to store character counts while preserving insertion order. Then I iterated through the map to find the first character with count = 1. ================================================== // Online Java Compiler // Use this editor to write, compile and run your Java code online import java.util.*; class Main {   public static void main(String[] args) {           String s="swiss";           char[] words=s.toCharArray();           Map<Character,Integer>map=new LinkedHashMap<Character,Integer>();           for(char word:words)     {      map.put(word,map.getOrDefault(word,0)+1);     }           for(Map.Entry<Character,Integer>entry:map.entrySet())     {       if(entry.getValue()==1)       {         System.out.println("First non-repeated character in the string is:"+entry.getKey());         break;       }     }               } } Output:First non-repeated character in the string is:w This was a good exercise to understand: Character frequency counting Importance of insertion order using LinkedHashMap String traversal logic #AutomationTestEngineer #Selenium #Java #CodingPractice #ProblemSolving

🚀 Day 6 of Java Series — Count Vowels Using Streams Ever wondered how to count vowels in a string using Java 8 in a clean and functional way? Here’s a simple yet powerful approach using Streams 👇 import java.util.*; import java.util.function.Function; import java.util.stream.Collectors; public class CountOfVowels { public static void main(String[] args) { String name = "Microservices"; List<String> vowels = Arrays.asList("a", "e", "i", "o", "u"); Map<String, Long> map = Arrays.stream(name.split("")) .collect(Collectors.groupingBy(Function.identity(), Collectors.counting())); List<Map.Entry<String, Long>> finalMap = map.entrySet().stream() .filter(entry -> vowels.contains(entry.getKey())) .toList(); System.out.println(finalMap); } } 🔍 How it works: 1️⃣ name.split("") → Converts string into individual characters 2️⃣ groupingBy(Function.identity(), counting()) → Counts frequency of each character 3️⃣ Filter step → Keeps only vowels 4️⃣ Final result → List of vowels with their count 👉 Output: [e=2, i=2, o=1] #Java #Java8 #Streams #Coding #Developers #Learning

Day 9 of Java Series ☕💻 Today’s topic is BufferedReader — a powerful way to take fast input in Java 🚀 🧠 What is BufferedReader? BufferedReader is a class in Java used to read text efficiently from input streams (like keyboard or file). It reads data in chunks (buffer) instead of character-by-character → faster performance ⚡ ⚙️ Why Use BufferedReader? ✔ Faster than Scanner ✔ Efficient for large input ✔ Reduces I/O operations ✔ Used in competitive programming 🔗 How it Works? 👉 Works with InputStreamReader to convert bytes into characters System.in → InputStreamReader → BufferedReader 💻 Example Code: import java.io.*; public class Main { public static void main(String[] args) throws Exception { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); System.out.print("Enter your name: "); String name = br.readLine(); System.out.println("Hello " + name); } } ⚡ Important Methods: readLine() → Reads full line read() → Reads single character close() → Closes stream ⚠️ Important Notes: Does NOT parse input automatically Must handle exceptions (IOException) Needs conversion for numbers 👉 Example: Java id="br2" Copy code int num = Integer.parseInt(br.readLine()); 🎯 Why Important? ✔ Used in interviews & CP ✔ Improves performance ✔ Important for backend & file handling 🚀 Pro Tip: Use BufferedReader + StringTokenizer for ultra-fast input in competitive programming 🔥 📢 Hashtags: #Java #BufferedReader #JavaSeries #Coding #Programming #Developers #LearnJava #Tech

Day14 Java Practice: Maximum Product of Three Elements in an Array While practicing Java, I solved an interesting array problem: 👉 Find the maximum product that can be formed using any three elements from the array. Example: Input: {10, 3, 5, 6, -20} At first, it looks like we just need the three largest numbers. But the twist is: negative numbers can change the result! 🧠 Key Idea: The product of two negative numbers becomes positive So we must compare: Product of the three largest numbers Product of two smallest (most negative) numbers and the largest number ================================================= // Online Java Compiler // Use this editor to write, compile and run your Java code online import java.util.*; class Main {   public static void main(String[] args) {     int a [] ={10,3,5,6,-20};     Arrays.sort(a);     int n=a.length;     System.out.println(Arrays.toString(a));     int result1=a[n-1]*a[n-2]*a[n-3];          int result2=a[0]*a[1]*a[n-1];          int result =Math.max(result1,result2);     System.out.println(result);        } } Output:[-20, 3, 5, 6, 10] 300 #JavaDeveloper #Arrays #CodingPractice #QualityEngineering #TechLearning

🚀 Java Streams Example: Extract First Letter of Each Word Given a string: String input = "Hello Java Program"; Output: HJP Java Stream Solution: import java.util.Arrays; public class FirstLetterOfWords { public static void main(String[] args) { String input = "Hello Java Program"; String result = Arrays.stream(input.split("\\s+")) .map(word->word.substring(0,1)) .collect(Collectors.joining()); System.out.println(result); } } Output: HJP Using Java Streams makes string manipulation clean and readable. #Java #JavaStreams #Coding #Programming #Developers #BackendDevelopment

@programiz:- Print words starting with vowel:- // Online Java Compiler // Use this editor to write, compile and run your Java code online import java.util.Scanner; class Main {   public static void main(String[] args) {          //print words starting(first word) with vowel     //words means more than 1 word means sentence     Scanner sc = new Scanner(System.in);     System.out.println("enter the sentence");     String sentence = sc.nextLine();    //  System.out.println(sentence);           String [] words = sentence.split(" ");     for(String word : words){       if(word.length()>0){         //checking the first char of word       char ch = Character.toLowerCase(word.charAt(0));       if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u'){         System.out.println(word);       }              }     }    }   }

💡 Java Interview Question How do you find the common elements from three lists in Java? Here’s a simple example: ✅ Two approaches: Using retainAll() with Set Using Java 8 Streams public class CommonElementFrom3List { public static void main(String[] args){ List<Integer> list1 = Arrays.asList(1, 2, 3, 4, 5); List<Integer> list2 = Arrays.asList(3, 4, 5, 6, 7); List<Integer> list3 = Arrays.asList(5, 6, 7, 8, 3); Set<Integer> common = new HashSet<>(list1); common.retainAll(list2); common.retainAll(list3); System.out.println(common); List<Integer> list = list1.stream() .filter(list2::contains) .filter(list3::contains) .distinct() .collect(Collectors.toList()); System.out.println(list); } } 📌 Output: [3, 5] ❓ Question for you: Which approach would you prefer in a real-world scenario and why? Also, how would you handle duplicate elements efficiently? #Java #CodingInterview #JavaDeveloper #Programming #TechLearning

🧠 Java Interview Question 👉 How to find duplicate characters in a String? Example: Input: "programming" Output: g, r, m Simple approach using HashMap: import java.util.HashMap; public class DuplicateCharacter { public static void main(String[] args) { String str = "programming"; HashMap<Character, Integer> map = new HashMap<>(); for (char c : str.toCharArray()) { map.put(c, map.getOrDefault(c, 0) + 1); } for (char c : map.keySet()) { if (map.get(c) > 1) { System.out.println(c); } } } }    💡 Logic: Count frequency of each character and print duplicates. 👉 Have you solved this in a different way? #Java #SpringBoot #Coding #InterviewQuestions #BackendDeveloper

🚀 Why is String Immutable but StringBuffer Mutable in Java? This is one of the most common and important interview questions for Java developers. 🔹 String (Immutable) Once created, it cannot be changed Every modification creates a new object Ensures security, thread-safety, and caching Used in sensitive areas like URLs, file paths, etc. 🔹 StringBuffer (Mutable) Can be modified after creation Changes happen in the same object More memory efficient Thread-safe (synchronized) 💡 Key Insight: Use String when data should not change Use StringBuffer when frequent modifications are needed #Java #JavaDeveloper #CoreJava #String #StringBuffer #Programming #Coding #SoftwareDevelopment #BackendDeveloper #FullStackDeveloper #SpringBoot #CodingInterview #InterviewPreparation #TechInterview #Developers #LearnJava #JavaConcepts #DSA #CodingLife #TechCommunity

Multithreading in Java refers to the capability of a program to execute multiple threads concurrently, enhancing performance and responsiveness. Ways to create a thread include: 1. **Extending the Thread class** ```java class MyThread extends Thread { public void run() { System.out.println("Thread is running"); } } ``` 2. **Implementing the Runnable interface** ```java class MyRunnable implements Runnable { public void run() { System.out.println("Thread is running"); } } ``` The thread lifecycle consists of the following states: - NEW: Thread created - RUNNABLE: Ready to run - RUNNING: Executing - TERMINATED: Execution completed Important thread methods include: - `start()`: Starts the thread - `sleep(ms)`: Pauses the thread - `join()`: Waits for another thread - `isAlive()`: Checks if the thread is active Synchronization is essential for controlling access to shared resources and preventing data inconsistency: ```java synchronized(this) { // critical section } ``` In summary: - Multithreading enhances performance. - Threads can be created using either the Thread class or the Runnable interface. - Synchronization is key for thread safety. As an interview tip, consider using ExecutorService (thread pools) in real-world applications instead of manually creating threads. Follow this series for 30 Days of Java Interview Questions. #java #javadeveloper #multithreading #codinginterview #backenddeveloper #softwareengineer #programming #developers #tech