Fibonacci Number LeetCode 509 Java Dynamic Programming

Day 10.2 of Java with DSA Journey 🚀 📌 Problem: Base 7 Conversion (LeetCode 504) At first, this looks like a simple math problem… But it actually teaches something deeper: 👉 How numbers are built in different systems 💡 Core Idea To convert a number from base 10 → base 7: Divide by 7 Store remainder Repeat until 0 Reverse the result 🧠 Key Learnings 🔹 Understanding Number Systems Decimal (base 10) is just one system—this logic works for ANY base 🔹 Modulo + Division Pattern num % base → current digit num / base → move to next position 🔹 Why Reverse? We generate digits from least significant → most significant ⚡ Complexity ⏱ Time: O(log₇ n) 📦 Space: O(log₇ n) 🔥 Pro Tips (Interview Level) 💡 Tip 1: This is a Universal Pattern This exact logic works for: Binary (base 2) Octal (base 8) Hex (base 16) 👉 Learn once, apply everywhere. 💡 Tip 2: Handle Negatives Separately Always convert using abs(num) and attach - later This avoids tricky modulo behavior. 💡 Tip 3: Avoid String Concatenation in Loops Use StringBuilder 👉 Shows awareness of time complexity + memory efficiency 💡 Tip 4: Reverse Thinking = Interview Gold If you're building output backward → think reverse at the end 💡 Tip 5: Hidden Optimization Insight If allowed, you could also: Use recursion → cleaner logic Or pre-calculate size for optimization 🔥 Real Insight This problem is not about Base 7… It’s about understanding: 👉 How computers represent and transform numbers internally Once you master this, you can convert between any number systems. Consistency builds depth 📈 #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Algorithms #MathLogic #Day10 #BitManipulation #InterviewPrep #CleanCode #Array #Optimization #MCA #lnct #100DaysOfCode #SoftwareEngineering #Algorithms #InPlaceAlgorithms #TechLearning #JavaDeveloper

Day 10 of Java DSA Journey 🚀 📌 Problem: Power of Two (LeetCode 231) Most people solve this using loops: ➡️ Keep dividing by 2 until you reach 1 But today I learned something better: 👉 You can solve it in ONE line using bit manipulation 🤯 💡 Core Idea A number is a power of two if: ✔️ It has only one ‘1’ in its binary form Examples: 1 → 0001 2 → 0010 4 → 0100 16 → 10000 🧠 The Magic Trick 👉 n & (n - 1) This removes the rightmost set bit So: If result = 0 → only one bit was set → ✅ Power of Two Else → ❌ Not a power of two ⚡ Complexity ⏱ Time: O(1) 📦 Space: O(1) 🔥 Pro Tips (Interview Level) 💡 Tip 1: Always Check Positivity First n > 0 is mandatory — bit tricks fail for negative numbers. 💡 Tip 2: Understand, Don’t Memorize n - 1 flips: the rightmost 1 → 0 all bits after it → 1 That’s why the trick works. 💡 Tip 3: This Pattern Appears Everywhere The same trick is used in: Counting set bits Subset generation Low-level optimizations 👉 Learn it once, reuse forever. 💡 Tip 4: Alternative Trick (Even Cleaner) Another way: 👉 (n & -n) == n This isolates the lowest set bit. If it's equal to n, only one bit exists. 💡 Tip 5: Bitwise = Senior-Level Thinking Using bit manipulation shows: ✔️ You understand how data is stored ✔️ You can optimize beyond brute force 🔥 Real Insight This problem is not about checking powers… It’s about recognizing: 👉 Patterns in binary representation Once you see that, the solution becomes obvious. Consistency builds mastery 🔑 #DSA #LeetCode #Java #BitManipulation #CodingJourney #ProblemSolving #InterviewPrep #Day10 #BitManipulation #InterviewPrep #CleanCode #Array #Optimization #MCA #lnct #100DaysOfCode #SoftwareEngineering #Algorithms #InPlaceAlgorithms #TechLearning #JavaDeveloper

Leetcode Practice - 15. 3Sum The problem is solved using JAVA. Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.   Example 1: Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter. Example 2: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0. Example 3: Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.   Constraints: 3 <= nums.length <= 3000 -105 <= nums[i] <= 105 #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

Today I explored some fundamental yet powerful concepts in Java that every developer should have a strong grip on: 🔹 Static Methods & VariablesUnderstanding how static members are shared across all objects really changed how I think about memory and efficiency. It’s amazing how a simple static keyword can help track object creation and maintain shared data seamlessly. 🔹 Constructor Overloading & this KeywordThis concept made object initialization much more flexible. Using multiple constructors and the this keyword not only improves code readability but also avoids redundancy. 💡 What I realized:Strong basics are the real game-changer. These concepts might look simple, but they build the foundation for writing clean, scalable, and efficient code. 📌 Consistency in learning > Complexity in topics I’m currently focusing on strengthening my core Java skills and building projects around them. Every small concept learned today contributes to becoming a better developer tomorrow. #Java #Programming #CodingJourney #DeveloperLife #JavaDeveloper #Learning #TechSkills #Coding #StudentDeveloper

🚀 Day 20 of #100DaysOfCode – Java DSA Journey Today was all about understanding some of the most important Java Collections — the building blocks for writing efficient code 💡 📚 Topic: ArrayList vs HashSet vs HashMap At first glance, they may look similar… but each serves a completely different purpose 👇 🔹 ArrayList ✔️ Ordered (maintains insertion order) ✔️ Allows duplicates ✔️ Index-based access 🧠 When to use? When order matters When you need to access elements using index When duplicates are allowed 🔹 HashSet ✔️ Unordered ✔️ No duplicates allowed ✔️ Faster lookups (O(1) average) 🧠 When to use? When you only care about unique elements When checking existence is important 🔹 HashMap ✔️ Stores data in key-value pairs ✔️ Keys are unique, values can be duplicated ✔️ Very fast operations (O(1) average) 🧠 When to use? When mapping relationships (like frequency count, indexing, caching) When you need quick access using keys 💭 Key Insight: Choosing the right data structure = cleaner code + better performance ⚡ Today made me realize: Not every problem needs a loop Sometimes, the right collection can reduce complexity instantly 📌 What I Learned Today: ✅ Difference between ArrayList, HashSet, and HashMap ✅ When to use each data structure ✅ Importance of avoiding duplicates efficiently ✅ Writing optimized logic using collections Consistency check ✅ Clarity improved ✅ Confidence growing 📈 Let’s keep building 🚀 Day 21 coming soon! #Java #DSA #100DaysOfCode #CodingJourney #LearningInPublic #DeveloperLife #Programmer #CodingLife #SoftwareEngineering #ComputerScience #TechJourney #ProblemSolving #Algorithms #DataStructures #JavaDeveloper #CodeDaily #Consistency #GrowthMindset #SelfImprovement #StudentLife #EngineeringStudent #FutureEngineer #CodeNewbie #KeepLearning #BuildInPublic #Motivation #Discipline #DailyProgress #NeverGiveUp

💡 Java Tip: One Method to Remember for Type Conversion We used to rely on: Integer.parseInt(), Long.parseLong(), Double.parseDouble() → for converting String to primitives String.valueOf() → for converting values to String It works—but it can get confusing when switching between primitives, wrapper classes, and even char ↔ String conversions. 🔑 Simple takeaway: You can simplify most conversions by remembering just one method: 👉 WrapperClass.valueOf() ✅ Converts String → Wrapper (Integer, Long, Double, etc.) ✅ Works well with primitives (via autoboxing/unboxing) ✅ Keeps your code more consistent and readable Example: Integer i = Integer.valueOf("10"); Double d = Double.valueOf("10.5"); String s = String.valueOf(100); 🧠 Personal learning: Instead of memorizing multiple parsing methods, focusing on valueOf() makes type conversion easier to reason about and reduces cognitive load while coding. #Java #CleanCode #ProgrammingTips #BackendDevelopment #SoftwareEngineering #Learning

Day 9.2 of Java with DSA Journey 🚀 📌 Problem: Number of Steps to Reduce a Number to Zero (LeetCode 1342) At first glance, this looks too easy… But hidden inside is a powerful idea: 👉 Thinking in binary instead of decimal 💡 Core Idea Keep reducing the number until it becomes 0: Even → divide by 2 Odd → subtract 1 Simple rule. Powerful pattern. 🧠 Key Learnings 🔹 Iterative Thinking Used a loop to repeatedly transform the state 🔹 Decision Making per Step Even vs Odd → determines next move 🔹 Bitwise Insight num % 2 == 0 → (num & 1) == 0 num / 2 → num >> 1 ⚡ Complexity ⏱ Time: O(log n) 📦 Space: O(1) 🔥 Pro Tips (Interview Level) 💡 Tip 1: Think in Binary, Not Decimal Every division by 2 removes one bit → that's why complexity is logarithmic. 💡 Tip 2: Count Operations Without Simulation Steps = 👉 (Number of bits - 1) + (Number of 1s in binary) Example: 14 → 1110 Steps = (4 - 1) + 3 = 6 💡 Tip 3: Bitwise > Arithmetic (When Optimizing) Replace: % 2 → & 1 / 2 → >> 1 This shows low-level understanding. 💡 Tip 4: Pattern Recognition Matters This problem is not about loops… It’s about recognizing bit reduction patterns. 💡 Tip 5: Always Look for Hidden Math Even simple problems often have a mathematical shortcut behind them. 🔥 Real Insight This problem teaches a subtle shift: ❌ “Keep applying rules” ✅ “Understand what each operation does to the binary structure” That’s how you move from coding → engineering. Consistency compounds 📈 #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Day9 #BitManipulation #InterviewPrep #CleanCode #Array #Optimization #MCA #lnct #100DaysOfCode #SoftwareEngineering #Algorithms #InPlaceAlgorithms #TechLearning #JavaDeveloper

🚀 Day 55: The Power of Polymorphism in Java 🎭 Today was a deep dive into one of the most powerful concepts in Object-Oriented Programming: Polymorphism (Greek for "Many Forms"). It’s the ability of an object to take on different forms depending on the context.   In Java, I learned that this flexibility happens at two distinct stages: 1. Compile-Time Polymorphism (Static Binding) ⏱️ This is achieved through Method Overloading.  ▫️ The Logic: Defining multiple methods in the same class with the same name but different parameters (type, number, or order). ▫️ The Benefit: It improves code readability and allows us to perform similar operations with different types of data without inventing new method names. Why "Compile-Time"? The compiler knows exactly which method to call just by looking at the arguments you provide. 2. Runtime Polymorphism (Dynamic Binding) 🏃♂️ This is achieved through Method Overriding.  ▫️ The Logic: When a subclass provides a specific implementation of a method that is already defined in its parent class. ▫️ The Magic: We use Upcasting (Parent class reference pointing to a Child class object). The specific version of the method to be executed is determined while the program is actually running. ▫️ The Benefit: This is the secret to building flexible, scalable systems where you can add new features without breaking existing code. Question for the Java Community: In your experience, what’s a real-world scenario where Runtime Polymorphism saved you from writing massive if-else or switch blocks? I’d love to hear your examples! 👇 #Java #OOPs #Polymorphism #100DaysOfCode #BackendDevelopment #CleanCode #SoftwareEngineering #LearningInPublic #JavaDeveloper 10000 Coders Meghana M

Day 3 of Java with DSA Journey 🚀 📌 Topic: Guess Number Higher or Lower (LeetCode 374) 💬 Quote: "Efficiency is not about doing more; it's about eliminating what doesn't matter." ✨ What I Learned: 🔹 Binary Search Beyond Arrays: Binary Search isn’t limited to arrays — it works perfectly on a number range like [1...n]. 🔹 Working with APIs: Learned how to adapt logic based on API responses: -1 → Guess is too high 1 → Guess is too low 0 → Correct answer 🔹 Power of Efficiency: Even for a huge range (up to 2³¹ - 1), Binary Search finds the answer in ~31 steps 🤯 Compared to Linear Search → practically impossible! 🔹 Complexity: ⏱ Time: O(log n) 📦 Space: O(1) 🧠 Problem Solved: ✔️ Guess Number Higher or Lower 💡 Key Insight: This problem highlights the “Narrowing the Search Space” concept. Each step eliminates half the possibilities — that’s the magic of logarithmic algorithms ⚡ ⚡ Interview Insight (3-Way Decision Logic): Unlike boundary problems, here we deal with three outcomes: 1️⃣ 0 → Found the number (return immediately) 2️⃣ -1 → Move right = mid - 1 3️⃣ 1 → Move left = mid + 1 👉 Use while (left <= right) since the target is guaranteed to exist. 🔑 Takeaway: Consistency beats intensity. Showing up daily is what builds mastery. #DSA #LeetCode #Java #CodingJourney #BinarySearch #ProblemSolving #100DaysOfCode #JavaDeveloper #Algorithms

🚀 LeetCode #496 – Next Greater Element I | Java Solution Today’s problem was a great introduction to Stacks + HashMaps — a powerful combination for optimizing search problems 🔥 📌 Problem Summary Given two arrays nums1 and nums2 (where nums1 is a subset of nums2), find the next greater element for each element in nums1 from nums2. 👉 The next greater element is the first element to the right that is larger than the current number. 💡 Approach I Used (Optimized – O(n)) Traverse nums2 using a monotonic decreasing stack For each element: If current element > stack top → it's the "next greater" Store mapping in a HashMap Finally, build result for nums1 using the map 💻 Java Code Snippet int[] arr = new int[n1.length]; Stack<Integer> stk = new Stack<>(); HashMap<Integer,Integer> map = new HashMap<>(); for(int i = 0; i < n2.length; i++){ while(!stk.isEmpty() && n2[i] > stk.peek()){ map.put(stk.pop(), n2[i]); } stk.push(n2[i]); } for(int j = 0; j < n1.length; j++){ arr[j] = map.getOrDefault(n1[j], -1); } 🧠 Key Learning Understanding Monotonic Stack Reducing brute force O(n²) → O(n) Using HashMap for quick lookups 📊 Example nums1 = [4,1,2] nums2 = [1,3,4,2] Output = [-1,3,-1] 🔥 Why this problem matters? Frequently asked in interviews Builds foundation for problems like: Daily Temperatures Stock Span Next Greater Element II Happy Coding 😊 #LeetCode #Java #DSA #Stack #CodingJourney #ProblemSolving #100DaysOfCode #Algorithms