Exception Handling in Java for Safer Code

🚀 Day 16 of Java Practice – First Repeating Element in Array Continuing my Java learning journey, today I solved an important array problem frequently asked in interviews. 💡 Problem: Find the first repeating element in an array 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { int arr[] = {10, 5, 3, 4, 3, 5, 6}; HashSet<Integer> set = new HashSet<>(); int firstRepeat = -1; for(int i = arr.length - 1; i >= 0; i--) { if(set.contains(arr[i])) { firstRepeat = arr[i]; } else { set.add(arr[i]); } } if(firstRepeat != -1) { System.out.println("First repeating element: " + firstRepeat); } else { System.out.println("No repeating element found"); } } } ✅ Output: 5 📌 What I learned: ✔ Using HashSet for duplicate detection ✔ Reverse traversal technique ✔ Optimized approach for array problems Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day16 #Freshers #OpenToWork

🚀 Day 18 of Java Practice – Intersection of Two Arrays Continuing my Java learning journey, today I solved an important array problem frequently asked in interviews. 💡 Problem: Find the intersection of two arrays 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { int arr1[] = {1, 2, 3, 4}; int arr2[] = {3, 4, 5, 6}; HashSet<Integer> set = new HashSet<>(); HashSet<Integer> result = new HashSet<>(); for(int num : arr1) { set.add(num); } for(int num : arr2) { if(set.contains(num)) { result.add(num); } } System.out.println("Intersection: " + result); } } ✅ Output: [3, 4] 📌 What I learned: ✔ Using HashSet for efficient lookup ✔ Finding common elements ✔ Avoiding duplicates automatically Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day18 #Freshers #OpenToWork

🚀 Day 12 of Java Practice – Find Duplicate Characters in String Continuing my Java learning journey, today I solved an important string problem frequently asked in interviews. 💡 Problem: Find duplicate characters in a string 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { String str = "programming"; HashMap<Character, Integer> map = new HashMap<>(); for(int i = 0; i < str.length(); i++) { char ch = str.charAt(i); map.put(ch, map.getOrDefault(ch, 0) + 1); } System.out.print("Duplicate characters: "); for(char ch : map.keySet()) { if(map.get(ch) > 1) { System.out.print(ch + " "); } } } } ✅ Output: r g m 📌 What I learned: ✔ HashMap for frequency counting ✔ Identifying duplicate elements ✔ Writing clean and efficient logic Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day12 #Freshers #OpenToWork

🚀 Day 10 of Java Practice – Check Anagram Strings Continuing my Java learning journey, today I solved a popular interview question. 💡 Problem: Check whether two strings are anagrams 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { String str1 = "listen"; String str2 = "silent"; str1 = str1.toLowerCase(); str2 = str2.toLowerCase(); char[] arr1 = str1.toCharArray(); char[] arr2 = str2.toCharArray(); Arrays.sort(arr1); Arrays.sort(arr2); if(Arrays.equals(arr1, arr2)) { System.out.println("Strings are Anagram"); } else { System.out.println("Strings are not Anagram"); } } } ✅ Output: Strings are Anagram 📌 What I learned: ✔ String to char array conversion ✔ Sorting arrays ✔ Comparing arrays Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day10 #Freshers #OpenToWork

🔥 Advanced Exception Handling in Java – Production-Grade Code Basic handling is not enough in real-world applications. To write clean, scalable, and production-ready code, you need advanced exception strategies. 🔹 1. Use Custom Exceptions Create your own exceptions to represent business logic clearly. class InvalidAmountException extends Exception { public InvalidAmountException(String msg) { super(msg); } } 🔹 2. Try-With-Resources (Auto Cleanup) No more manual closing of resources 👇 try (BufferedReader br = new BufferedReader(new FileReader("file.txt"))) { System.out.println(br.readLine()); } 🔹 3. Exception Chaining Preserve original cause for better debugging. throw new RuntimeException("Error processing", e); 🔹 4. Global Exception Handling (Spring Boot) Use @ControllerAdvice to handle exceptions centrally in REST APIs. 💡 Pro Tip: Log exceptions properly (using Logger) instead of just printing messages. 🚀 Advanced exception handling is what separates junior code from enterprise-level applications. #Java #ExceptionHandling #AdvancedJava #SpringBoot #BackendDevelopment #SoftwareEngineering #CleanCode #Coding #Developers #TechSkills #Programming #JavaDeveloper #FullStackDeveloper #OpenToWork #Hiring #JobSearch #Freshers #Opportunities #Careers #TechJobs #InterviewPreparation #CodeNewbie #LinkedInTech #100DaysOfCode #DeveloperLife #ITJobs

🚀 Day 6 of Java Practice – First Non-Repeating Character Continuing my Java learning journey, today I solved an important string problem. 💡 Problem: Find the first non-repeating character in a string 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { String str = "programming"; HashMap<Character, Integer> map = new HashMap<>(); for(int i = 0; i < str.length(); i++) { char ch = str.charAt(i); map.put(ch, map.getOrDefault(ch, 0) + 1); } for(int i = 0; i < str.length(); i++) { char ch = str.charAt(i); if(map.get(ch) == 1) { System.out.println("First non-repeating character: " + ch); break; } } } } ✅ Output: p 📌 What I learned: ✔ HashMap for frequency counting ✔ Problem-solving using 2-step logic ✔ Real interview question practice Improving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day7 #Freshers #OpenToWork

🚀 Day 19 of Java Practice – Longest Substring Without Repeating Characters Continuing my Java learning journey, today I solved a very important interview problem using the sliding window technique. 💡 Problem: Find the length of the longest substring without repeating characters 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { String str = "abcabcbb"; HashSet<Character> set = new HashSet<>(); int left = 0, maxLength = 0; for(int right = 0; right < str.length(); right++) { while(set.contains(str.charAt(right))) { set.remove(str.charAt(left)); left++; } set.add(str.charAt(right)); maxLength = Math.max(maxLength, right - left + 1); } System.out.println("Longest length: " + maxLength); } } ✅ Output: 3 📌 What I learned: ✔ Sliding Window technique ✔ Using HashSet for duplicate tracking ✔ Optimized string handling Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day19 #Freshers #OpenToWork

🚀 Day 22 of Java Practice – Second Largest Element in Array Continuing my Java learning journey, today I solved an important array problem frequently asked in interviews. 💡 Problem: Find the second largest element in an array 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { int arr[] = {10, 5, 20, 8, 15}; int first = Integer.MIN_VALUE; int second = Integer.MIN_VALUE; for(int i = 0; i < arr.length; i++) { if(arr[i] > first) { second = first; first = arr[i]; } else if(arr[i] > second && arr[i] != first) { second = arr[i]; } } System.out.println("Second largest: " + second); } } ✅ Output: Second largest: 15 📌 What I learned: ✔ Finding max & second max in single loop ✔ Optimized O(n) solution ✔ Handling edge cases Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day22 #Freshers #OpenToWork

🚀 Day 21 of Java Practice – Merge Two Arrays Continuing my Java learning journey, today I solved an important array problem. 💡 Problem: Merge two arrays into a single array 👉 Solution: import java.util.*; class MainClass { public static void main(String args[]) { int arr1[] = {1, 3, 5}; int arr2[] = {2, 4, 6}; int n1 = arr1.length; int n2 = arr2.length; int merged[] = new int[n1 + n2]; for(int i = 0; i < n1; i++) { merged[i] = arr1[i]; } for(int i = 0; i < n2; i++) { merged[n1 + i] = arr2[i]; } System.out.println(Arrays.toString(merged)); } } ✅ Output: [1, 3, 5, 2, 4, 6] 📌 What I learned: ✔ Array traversal ✔ Merging arrays ✔ Handling multiple arrays Improving problem-solving step by step 🚀 Open to opportunities as a Java Developer (Fresher). #Java #CodingJourney #Day21 #Freshers #OpenToWork