Power of Four LeetCode 342 Java Solution

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 11 of Java with DSA Journey — This one blew my mind 🤯 📌 Problem: Power of Three (LeetCode 326) Yesterday, I used bit manipulation for Power of Two. Today? 👉 That approach completely fails. So I had to think differently… 💡 Breakthrough Idea Instead of loops or recursion… I used math + number theory to solve it in O(1) time. 👉 1162261467 % n == 0 Yes, one line. 🧠 Key Learnings 🔹 Bitwise isn’t universal Works great for base-2, but not for base-3 🔹 Prime Numbers Matter Since 3 is prime, its powers divide each other perfectly 🔹 Max Value Trick Largest power of 3 in 32-bit int = 3^19 = 1162261467 ⚡ Complexity ⏱ Time: O(1) 📦 Space: O(1) 🔥 Pro Tips (Interview Level) 💡 Tip 1: Know Your Data Type Limits Many O(1) tricks depend on constraints like 32-bit integer limits 💡 Tip 2: Prime Numbers Unlock Shortcuts If a number is prime, its powers have clean divisibility properties 💡 Tip 3: Always Question the Default Approach Most people write: while (n % 3 == 0) n /= 3; 💡 Tip 4: This is a Pattern Power of 2 → bitwise Power of 3 → math Power of 4 → hybrid 💡 Tip 5: Edge Cases First Always check n <= 0 🔥 Real Insight This problem forced me to shift my thinking: ❌ Rely on one technique ✅ Adapt based on the problem Consistency compounds 📈 #DSA #LeetCode #Java #CodingJourney #ProblemSolving #NumberTheory #InterviewPrep #Day11 #BitManipulation #CleanCode #Array #Optimization #MCA #lnct #100DaysOfCode #SoftwareEngineering #Algorithms #InPlaceAlgorithms #TechLearning #JavaDeveloper

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

Day 4 of Java with DSA Journey 🚀 📌 Topic: Search Insert Position (LeetCode 35) 💬 Quote: "Binary Search is about more than finding a needle in a haystack; it's about knowing exactly where to put a new needle." ✨ What I Learned: 🔹 Power of the low Pointer: If the target isn’t found, low directly gives the correct insertion index. 🔹 Finding Position Even When Missing: Binary Search doesn’t just search — it tells you where the element belongs. 🔹 Efficient Gap Detection: Even for missing values, we maintain O(log n) efficiency. 🔹 Complexity: ⏱ Time: O(log n) 📦 Space: O(1) 🧠 Problem Solved: ✔️ Search Insert Position 💡 Key Insight: Binary Search helps determine the rank/position of a number in a sorted array — whether it exists or not ⚡ ⚡ Interview Insight (Post-Loop Behavior): 👉 When the loop ends (low > high): low → first index greater than target high → last index smaller than target 🎯 That’s why low = insert position 🔑 Takeaway: Binary Search is not just about finding — it's about positioning. Consistency is the real key 🔑 #DSA #LeetCode #Java #CodingJourney #BinarySearch #ProblemSolving #100DaysOfCode #Algorithms #TechLearning #Day4 #Array #MCA #lnct

🚀 Learning Java the Right Way Today, I practiced a popular DSA problem 👉 Container With Most Water 📌 Problem: Given an array of heights, find two lines that together can store the maximum amount of water. Example: Array → {1, 8, 6, 2, 5, 4, 8, 3, 7} Output → 49 ✅ 🔹 Key Learning: Instead of using brute force (O(n²)), I used the Two Pointer approach to solve it in O(n) time. 📌 Approach: Start with two pointers (left & right) Calculate area = min(height) × width Move the pointer with smaller height Repeat to find maximum area This problem helped me understand: ✔ Two Pointer technique ✔ Optimization over brute-force ✔ Logical decision making ✔ Real interview-level problem solving Learning when and how to optimize is what truly improves coding skills 💪 📌 Think smart • Optimize logic • Solve efficiently 🚀 #java #javafullstack #javadeveloper #corejava #codingjourney #coding

Day 7 of Java with DSA Journey 🚀 📌 Topic: Missing Number (LeetCode 268) Quote: "Sometimes the best way to solve a coding problem is to put down the keyboard and pick up a calculator." ✨ What I learned today: 🔹 Mathematical Optimization: Instead of brute force, I used a simple formula to compute the expected sum. {Sum} = \frac{n(n+1)}{2} 🔹 Core Idea: Expected Sum − Actual Sum = Missing Number 🔹 Alternative Approach: Used XOR technique where duplicates cancel out → leaving the missing value. 🔹 Efficiency: ✔️ Time Complexity: O(n) ✔️ Space Complexity: O(1) 🧠 Problem Solved: ✔️ Missing Number 💡 Key Insight: We often jump to HashSet or Sorting, but this problem taught me to pause and think mathematically first. A simple formula reduced space complexity from O(n) → O(1) 🚀 ⚡ Interview Tip (Important!): Be careful of integer overflow in Java: n * (n + 1) can exceed int limit Use long OR switch to XOR method for safety Consistency is the real key 🔑 #DSA #LeetCode #Java #Mathematics #CodingJourney #ProblemSolving #Day7 #Algorithms #Optimization #Array #MCA #lnct #100DaysOfCode #Algorithms #SoftwareEngineering #InPlaceAlgorithms #TechLearning #JavaDeveloper

Day 6 of Java with DSA Journey 🚀 📌 Topic: Max Consecutive Ones (LeetCode 485) 💭 Quote: "Progress is about maintaining momentum—just like a streak of consecutive ones." ✨ What I learned today: 🔹 State Tracking: Learned how to maintain a current streak vs a maximum streak — a key concept for sequence-based problems. 🔹 The Power of Resetting: Understanding when to reset the counter (on encountering 0) is crucial to solving streak problems correctly. 🔹 Simple Yet Powerful Logic: No complex data structures needed — just a clean linear scan with strong logic. 🔹 Performance: ✔️ Time Complexity: O(n) (Single pass) ✔️ Space Complexity: O(1) (Only variables used) 🧠 Problem Solved: ✔️ Max Consecutive Ones 💡 Key Insight: This problem is a foundation for the Sliding Window pattern. Instead of managing two pointers, we track a running window of 1s that resets when a 0 appears. ⚡ Interview Insight – Using Math.max(): Instead of writing: if (current > max) max = current; Using: max = Math.max(max, current); makes the code cleaner and more professional — a small detail that reflects strong coding practices. 📈 Growth Reflection: With each problem, I’m focusing on: 🔹 Explaining logic clearly and step-by-step (important for interviews) 🔹 Thinking in patterns (like Sliding Window & State Tracking) 🔹 Writing clean and optimized code like a software engineer Consistency is the real key 🔑 #DSA #LeetCode #Java #CodingJourney #SlidingWindow #ProblemSolving #Day6 #SoftwareEngineer #Algorithms #Array #MCA #lnct #100DaysOfCode #Algorithms #SoftwareEngineering #InPlaceAlgorithms #TechLearning #JavaDeveloper

🚀 Day 18 / 100 Days of Java 💻 Today was all about going deeper into one of the most fundamental topics in programming — Arrays. Even though arrays seem simple at first, they are the backbone of many advanced data structures and algorithms. Here’s what I worked on today 👇 🔹 Understanding Arrays Learned how arrays are represented in memory and why they allow fast access using indexing. This helped me clearly understand how data is stored and retrieved efficiently. 🔹 Finding Maximum & Minimum Elements Practiced iterating through an array to identify the largest and smallest values — a simple yet powerful concept used in many real-world problems. 🔹 Finding the Third Largest Element This pushed me to think beyond basics and handle edge cases like duplicates and ordering without relying completely on sorting. 🔹 Searching an Element in Array Explored linear search and understood where it works best. Also got a glimpse of how search efficiency matters when data grows. 🔹 Finding Missing Number Solved problems using both brute force and optimized approaches. This improved my understanding of patterns and mathematical logic. 🔹 Finding Repeating Elements Learned different techniques to detect duplicates — from basic loops to more optimized methods using extra space. 💡 Key Learnings from Today: ✔ Arrays are not just beginner topics — they are the foundation of problem solving ✔ Writing clean logic is more important than jumping to complex solutions ✔ Edge cases (duplicates, boundaries, etc.) matter a lot ✔ There’s always a better (optimized) way to solve a problem 🔥 Reflection: Every day I realize that consistency beats intensity. Even small concepts, when practiced deeply, build strong problem-solving skills over time. 📈 Slowly but surely becoming better than yesterday. Let’s keep building, learning, and growing 💪 #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

🚀 𝐍𝐞𝐰 𝐕𝐢𝐝𝐞𝐨 𝐎𝐮𝐭 — 𝐖𝐡𝐲 𝐀𝐫𝐫𝐚𝐲 𝐈𝐧𝐝𝐞𝐱 𝐒𝐭𝐚𝐫𝐭𝐬 𝐟𝐫𝐨𝐦 𝟎? (Beginner Friendly) Ever wondered why arrays in Java start from 0 instead of 1? 🤔 This is one of the most common questions beginners have — and understanding this will level up your programming fundamentals. In this video, I’ve explained the concept in a simple and practical way — not just theory, but the real reason behind it. 🎬 𝐓𝐨𝐩𝐢𝐜: Why Index Starts at 0 in Arrays 💡 𝐈𝐧 𝐭𝐡𝐢𝐬 𝐯𝐢𝐝𝐞𝐨 𝐲𝐨𝐮’𝐥𝐥 𝐥𝐞𝐚𝐫𝐧: ✅ What is an Array & what is Index ✅ Why indexing starts from 0 (core concept) ✅ Memory concept behind arrays (important 🔥) ✅ How JVM calculates element position ✅ Formula: base_address + (index * size) ✅ Why starting from 0 makes access faster 💡 I’ve explained this using a simple memory visualization: • First element → base address (index 0) • Next elements → calculated using offset • No extra calculation needed → more efficient ⚡ This helps you understand how things work internally, not just remember syntax. 📺 Watch Video 👇 https://lnkd.in/ga3iaUNN 💬 Did you know this before, or did you think arrays should start from 1? 🔁 Share this with someone learning Java — this concept clears a lot of confusion 🚀 #Java #JavaCourse #LearnJava #Programming #SoftwareEngineering #Developers #Coding #Arrays #DataStructures #CodingJourney #NVerse

🚀 DAY 99/150 — BINARY SEARCH ON ANSWER! 🔥📊 Day 99 of my 150 Days DSA Challenge in Java and today I solved a very important problem that uses Binary Search in a 2D matrix 💻🧠 📌 Problem Solved: Kth Smallest Element in a Sorted Matrix 📌 LeetCode: #378 📌 Difficulty: Medium–Hard 🔹 Problem Insight The task is to find the kth smallest element in an n x n matrix where: • Each row is sorted • Each column is sorted 👉 Instead of flattening and sorting (O(n² log n)), we can do better using Binary Search on Answer. 🔹 Approach Used I used Binary Search on Value Range: • Set: low = smallest element high = largest element • Apply binary search: Find mid Count how many elements are ≤ mid 👉 If count < k → move right (low = mid + 1) 👉 Else → move left (high = mid) ✔️ Continue until low == high → answer found 🔹 Counting Logic (Optimized) • Start from bottom-left corner • If element ≤ mid: All elements above are also ≤ mid Move right • Else: Move up ✔️ This gives O(n) counting per iteration ⏱ Complexity Time Complexity: O(n log (max - min)) Space Complexity: O(1) 🧠 What I Learned • Binary Search is not just for arrays — it can be applied on answer space • Matrix properties can be used for efficient counting • Avoid brute force by leveraging sorted structure 💡 Key Takeaway This problem taught me: How to apply Binary Search on Answer pattern How to optimize 2D problems using structure Thinking beyond direct sorting approaches 🌱 Learning Insight Now I’m improving at: 👉 Recognizing when to use Binary Search beyond arrays 👉 Solving problems with optimized thinking 🚀 ✅ Day 99 completed 🚀 51 days to go 🔗 Java Solution on GitHub: 👉 https://lnkd.in/g-yhc7kH 💡 Don’t search the data — search the answer. #DSAChallenge #Java #LeetCode #BinarySearch #Matrix #Optimization #150DaysOfCode #ProblemSolving #CodingJourney #InterviewPrep #LearningInPublic