Java Nested Loop Control and Logic Fundamentals

🚀Day 20 of #120DaysOfCode 📌 Problem: Reverse Integer(7) 📌 Language: Java Algorithm: 1. Initialize rev = 0 2. While x != 0 . Extract digit . Check overflow . Update rev 3. Return rev ⏱️ Time and Space Complexity Time Complexity: O(logn) Space complexity: O(1) 🔥One problem closer to mastery #120DaysOfCode #Day20 #Java #Array #Leetcode #ProblemSolving #Consistency #LearningEveryday #LearningPublic #DSA

🚀Day 14 of #120DaysOfCode 📌 Problem: Matrix Reshape(566) 📌 Language: Java 🔍Approach 1. Check if Reshape is possible 2. Traverse the original matrix in row order 3. Mapping logic ⏱️ Time and Space Complexity Time Complexity: O(m x n) Space complexity: O(r x c) 🔥One problem closer to mastery #120DaysOfCode #Day14 #Java #Array #Leetcode #ProblemSolving #Consistency #LearningEveryday #LearningPublic #DSA

Today’s LeetCode POTD: Key takeaway: leveraging inorder traversal + divide-and-conquer can significantly improve both runtime and memory efficiency when balancing a BST. Small daily improvements compound over time. #LeetCodeDaily #DataStructures #Algorithms #Java #LearningInPublic

📅 Day 7 – Java Full Stack Development with AI Today I learned about static keyword and Class & Object in Java. Topics covered: static variables and static methods Why static does not need object creation Class and Object basics Key takeaway: static members belong to the class, and objects are used to access non-static members. #CoreJava #StaticKeyword #ClassAndObject #JavaLearning #FullStackDeveloper #Day7

DSA journey 🚀 📌 LeetCode #1 – Two Sum 💻 Language: Java 🔹 Approach: Use two nested loops to check all possible pairs Compare the sum of each pair with the target Return indices once the matching pair is found ⏱ Time Complexity: O(n²) 🧩 Space Complexity: O(1) Consistency over perfection 💯 #DSA #Java #LeetCode #ProblemSolving #LearningInPublic #DSAWithedSlash

Day 12 of #100DaysOfLeetCode 💻✅ Solved #326. Power of Three problem in Java. Approach: • Checked if the number is less than or equal to zero (not a power of three) • Continuously divided the number by 3 while it was divisible • Verified whether the final value becomes 1 • Avoided recursion and used iteration for better efficiency Performance: ✓ Runtime: 8 ms (Beats 86.60% submissions) ✓ Memory: 46.35 MB Key Learning: ✓ Strengthened understanding of number divisibility logic ✓ Learned efficient validation of power-based numbers ✓ Improved problem-solving using iterative reduction Learning one problem every day 🚀 #Java #LeetCode #DSA #ProblemSolving #CodingJourney #100DaysOfCode

DSA journey 🚀 📌 LeetCode #509 – Fibonacci Number 💻 Language: Java 🔹 Approach: - Handle base cases (n == 0 and n == 1) - Use an iterative approach instead of recursion - Build the Fibonacci sequence using variables ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(1) Consistency over perfection 💯 #DSA #Java #LeetCode #ProblemSolving #LearningInPublic

Loops running forever? Code executing when it shouldn’t? 😬 Meet jump statements. Part 3: Jump Statements (break, continue, return). Learn how to stop, skip, or exit — the right way. Clean flow → Clean code ✨. 👉 Follow for more Java fundamentals. 🔁 Save & repost to revise the full Control Statements series. . . Thanks to Harshita Mittal for the design touch! . . #Java #CoreJava #JavaProgramming #JavaBasics #ControlStatements #JumpStatements #Break #Continue #Return #LearnJava #ProgrammingConcepts #Coding #CodeNewbie #BeginnerFriendly #SoftwareEngineering #DeveloperJourney #JavaSeries #LinkedInLearning #LogicBuilding #CleanCode

Day 4 — Fundamentals Today → Relational & Logical Operators Conditional Statements (if–else, switch) Small problem-solving challenges. Big lesson: Writing conditions is easy. Writing clean logic is hard. Day 5 tomorrow. #LearnInPublic #Java #BuildInPublic #CodingJourney #BackendDevelopment

DSA journey 🚀 📌 LeetCode #167 – Two Sum II (Input Array Is Sorted) 💻 Language: Java 🔹 Approach (Two Pointer Technique): Initialize two pointers at the start and end of the sorted array Calculate the sum of both pointers If the sum equals the target → return 1-based indices If the sum is smaller than the target → move the start pointer forward If the sum is greater than the target → move the end pointer backward Efficient use of the sorted property 💡 ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(1) Step-by-step clarity before optimization ✨ Consistency over perfection 💯 #DSA #Java #LeetCode #ProblemSolving #LearningInPublic