Solved House Robber Problem in Java Using DP

Finally implemented the 0/1 Knapsack problem using Dynamic Programming in Java 🧠💻 While revisiting classic algorithms, I decided to go old school with Java — and this time, I went deep into the Knapsack problem. It’s fascinating how such a seemingly simple problem teaches you so much about: Optimal substructure Overlapping subproblems And most importantly, how dynamic programming can turn exponential recursion into an elegant bottom-up solution. Watching those arrays fill up and tracing back the solution always gives that satisfying “it finally clicked” moment. 😅 Sometimes, revisiting the basics is the best way to sharpen your fundamentals. Have you ever gone back to re-implement classic DSA problems after years of working in real-world projects? #Java #DynamicProgramming #Algorithms #DSA #ProblemSolving #Knapsack

👋 Hi Connections! 💻 Today I worked on an interesting Java problem — finding unique pairs in an array whose sum equals a target value! 🎯 🧩 Problem: Given an array and a target, find how many unique pairs exist such that their sum = target. ⚙️ My Approach: I used the two-pointer technique on a sorted array: ➡️ Initialize two pointers (start and end). ➡️ If the current sum is less than the target → move start forward. ➡️ If the sum is greater → move end backward. ➡️ If the sum equals the target → record it, skip duplicates, and move both pointers. ✨ This avoids unnecessary comparisons and handles duplicates smartly, giving a clean O(n) solution! 💡 What I Learned: ✅ The two-pointer method is perfect for pair-sum problems. ✅ Skipping duplicates carefully improves accuracy and avoids redundancy. ✅ Thinking in patterns rather than brute-force leads to elegant solutions. Every problem solved is one more step in improving my DSA intuition and optimization mindset 💪🔥 #Java #DSA #Coding #ProblemSolving #TwoPointers #LeetCode #Programming #LearningEveryday

🚀 Day 53 | DSA with Java Today, I solved the Aggressive Cows Problem — another classic example of Binary Search on Answers. 🐄🏠 --- 🔹 Problem Statement: Given n stalls at different positions and c cows: Place the cows in stalls so that the minimum distance between any two cows is maximized. Find the largest minimum distance possible. Example: Input: stalls = [1,2,4,8,9], c = 3 Output: 3 ✅ (Place cows at positions 1, 4, 8) --- ⚙️ Approach – Binary Search on Answers 1. Sort the stalls array. 2. Search in the distance range [1, max(stalls) - min(stalls)]. 3. For each mid distance, check if it’s possible to place all cows maintaining at least mid distance. 4. Adjust the search space based on feasibility: If possible → try larger distance Else → try smaller distance Time Complexity: O(n × log(maxDistance)) Space Complexity: O(1) --- 🧠 Learning Outcomes: ✅ Practiced Binary Search on Answers in a spatial context ✅ Strengthened logical thinking with feasibility functions ✅ Similar patterns as Book Allocation and Painters Partition --- #DSA #Java #BinarySearch #AggressiveCows #ProblemSolving #Coding #Programming #100DaysOfCode #GitHub #LogicBuilding

🔹 Mastering Inheritance in Java Inheritance allows one class to extend another class and reuse its properties and methods. It promotes code reusability and helps in writing cleaner, organized programs. // Parent Class (Super Class) class Animal { String type = "Animal"; void sound() { System.out.println("Animal makes a sound"); } } // Child Class (Sub Class) class Dog extends Animal { // Method Overriding @Override void sound() { System.out.println("Dog barks"); } } public class Main { public static void main(String[] args) { Dog d = new Dog(); System.out.println(d.type); // Inherited property d.sound(); // Calls overridden method } } ✅ Output: Animal Dog barks 🔍 Why It Matters: Reusability: Avoid rewriting the same code. Maintainability: Changes in one place reflect everywhere. Polymorphism: Child can modify parent behavior (as shown above). 🧠 Remember: Use inheritance only when there is a clear “is-a” relationship. #Java #OOP #Inheritance #ProgrammingFundamentals #CleanCode #CodeReusability #LearningInPublic #Developers #TechJourney

Today, I practiced an interesting concept in Java — "Rotating an array from left to right and right to left using the reversal algorithm." *CONCEPT OVERVIEW Array rotation means shifting array elements by a certain number of positions. For example: Right to left Rotation: [1, 2, 3, 4, 5] → [5, 1, 2, 3, 4] Left to right Rotation: [1, 2, 3, 4, 5] → [2, 3, 4, 5, 1] To achieve this efficiently, instead of using extra arrays or nested loops, I implemented a reversal-based approach, which works in-place (O(n) time and O(1) space). *Logic Used: Right Rotation Steps: --> Reverse the entire array --> Reverse the first r elements --> Reverse the remaining elements Left Rotation Steps: --> Reverse the entire array --> Reverse the first (n – r) elements --> Reverse the last r elements *CODE STRUCTURE : rotationalArray() -> handles the main rotation logic. reverseArray() -> reverses elements between two given indices. Used Arrays.toString() for easy printing of array content. *Edge Case: If the number of rotations exceeds the array length, an error message is displayed. #Java #Coding #Programming #LearningJourney#Arrays #ReversalAlgorithm #CodePractice

✨ Just Implemented LeetCode challenges — “Minimum Window Substring” using Java! This problem truly sharpened my understanding of the Sliding Window technique and taught me how efficient algorithms depend on balancing two moving pointers 🧠💻   Key Steps: Initialization: Create a frequency map (need) for characters in T. Initialize a counter (required) with T's length.      2.  Expansion (Right Pointer): The right pointer expands the window one character at a time.    3.Contraction (Left Pointer): Once required is 0 (a valid window is found), the inner while loop starts contracting the window from the left.    4. Result: After the entire string $S$ is traversed, return the minimum window substring found  ⚙️ Time Complexity: O(n) — Each character is processed at most twice (once by right and once by left). 🧠 Space Complexity: O(1) — Constant space, since the frequency array size is fixed (128 ASCII characters).   🔑 Takeaway: Efficient coding isn’t just about writing code that works — it’s about writing code that adapts smartly to changing conditions. 🖼️ Caption for Code Image “Sliding windows, shifting focus — finding clarity through logic 🪟💡” #LeetCode #Java #CodingChallenge #ProblemSolving #DataStructures #Algorithms #LearningJourney #SoftwareEngineering #WomenInTech

🚀 Day 91 of #100DaysOfCode 🧠 Solved "License Key Formatting" in Java using a reverse traversal + StringBuilder approach. 🔹 Problem: Given a license key string containing alphanumeric characters and dashes, reformat it so that each group (except possibly the first) has exactly k characters, dashes are placed between groups, and letters are uppercase. 🔹 Approach: • Traverse the input string **from end → start**, skip dashes.   • For each valid character: convert to uppercase, append to builder, and after every k characters, append a dash.   • Remove a trailing dash if one exists, then reverse the builder to get the final string. 🔹 Key Learning: • Building groups from the **backwards** simplifies handling of the “first group can be shorter” rule.   • `StringBuilder` avoids expensive string concatenations in loops.   • Good pattern for string-reformat problems: reverse + build + reverse. 🔹 Thanks to: • Dr Bharathi Raja N — Director of Training & Placement, Dhanalakshmi College of Engineering   • Sreesairam V Sir — Mentor from Aptitude Guru Hem #DhanalakshmiCollegeOfEngineering #LeetCode   #Java #100DaysOfCode #Day91 #StringManipulation #CodingJourney

𝑼𝒏𝒅𝒆𝒓 𝒕𝒉𝒆 𝑯𝒐𝒐𝒅: 𝑯𝒐𝒘 𝑷𝒓𝒐𝒈𝒓𝒂𝒎𝒎𝒊𝒏𝒈 𝑳𝒂𝒏𝒈𝒖𝒂𝒈𝒆𝒔 𝑹𝒆𝒂𝒍𝒍𝒚 𝑾𝒐𝒓𝒌 Have you ever wondered how our code actually talks to the computer? 🤔 Computers only understand binary (0s and 1s), but we write in languages like Java, Python, or JavaScript. So, how does our code get translated into something a machine understands? That’s where compilers and interpreters come in. Compiler → Translates the entire program into machine code before execution. Fast execution Errors shown after compilation Example: C, C++, Go Interpreter → Reads and executes line by line. Easy to debug Slower performance Example: Python, JavaScript Java is a mix of both! It first compiles code into bytecode, and then the JVM interprets it into machine code. That’s why Java is “Write Once, Run Anywhere.” JavaScript, on the other hand, runs directly in browsers using engines like V8, which now use Just-In-Time (JIT) compilation to boost speed — combining the best of both worlds. In short: Compilers prepare the whole meal before serving. Interpreters cook each bite as you eat! Which one do you prefer working with — compiled or interpreted languages? #Programming #Java #JavaScript #SoftwareDevelopment #Learning

🔄 Java Multithreading I’ve used thread.start() a hundred times — but never stopped to think what actually happens next. 🤔 Threads in Java go through a few states — and understanding them makes debugging so much easier. Here’s the quick flow 👇 NEW → When you create a thread object but haven’t started it yet. Thread t = new Thread(() -> {}); RUNNABLE → After calling start(). It’s ready to run, waiting for CPU time. BLOCKED / WAITING / TIMED_WAITING → When it’s paused — maybe waiting for a lock or sleeping. Thread.sleep(1000); TERMINATED → Once run() finishes, the thread’s life ends. Why does this matter? Because knowing where a thread is can help you spot issues like deadlocks, long waits, or threads that never end. Next time your code hangs, check its state — it often tells the full story. If you enjoyed this, follow me — I’m sharing one Java Multithreading concept every other day in simple language. And if you’ve ever debugged a “stuck” thread, share how you figured it out 💬 “Every concept you truly understand adds another layer to your confidence.” 🌱 #Java #Multithreading #ThreadLifecycle #Concurrency #BackendDevelopment #SpringBoot #Microservices #Coding #Learning

👇 🚀 Java Logic Series – 1: Find the Missing Number in an Array Today’s quick coding exercise — finding a missing number from a sequence using a simple mathematical approach. int[] array = {1, 2, 3, 4, 6, 7, 8, 9}; int n = array.length + 1; // total elements including missing one int expectedSum = n * (n + 1) / 2; int actualSum = 0; for (int num : array) { actualSum += num; } int missingNumber = expectedSum - actualSum; System.out.println("Missing number is: " + missingNumber); 🧠 Logic Behind It: ✅ Formula for sum of first n natural numbers → n*(n+1)/2 ✅ Subtract the actual sum of the array → get the missing number ✅ Time Complexity → O(n) 💡 Output: 👉 Missing number is: 5 Sometimes, the simplest math-based logic solves the problem elegantly. #Java #Coding #Programming #JavaDeveloper #LogicBuilding #ProblemSolving #InterviewPreparation #CleanCode