Solved Palindrome Number in Java with 2 pointers

🌳 Day 68 of Posting on LinkedIn LeetCode 112. Path Sum — Java Solution Today’s problem checks whether a root-to-leaf path exists in a binary tree such that the sum of the node values equals a target sum. This is a classic DFS (Depth-First Search) problem. We recursively explore left and right subtrees, subtracting the node’s value from the target at each step. When we reach a leaf node, we simply verify if the remaining sum matches the leaf’s value. ✅ Approach Use DFS recursion. At each node, subtract its value from the target sum. When a leaf node is reached, check if the remaining sum is zero. Return true if any path satisfies the condition. 🧠 Key Takeaways Classic DFS problem on binary trees Base case: checking leaf nodes Use subtraction to track required remaining sum #leetcode #leetcode112 #dsa #coding #java #binarytrees #programming #tech #interviewprep #softwareengineering #100daysofcode

Day 23 of #50DaysOfCode – Java 💻 Today’s challenge was to check whether a number is a Perfect Number. A Perfect Number is a number that is equal to the sum of all its positive divisors (excluding itself). Example: 6 → 1 + 2 + 3 = 6 28 → 1 + 2 + 4 + 7 + 14 = 28 This problem helped me strengthen my understanding of loops, conditions, and mathematical logic 🔍✨ #Java #CodingChallenge #50DaysOfCode #LearnToCode #ProgrammingBasics #LogicBuilding #CodeDaily #ProblemSolving

29/30 days ✅ Solved a LeetCode Problem #231 Power of Two — Bit Manipulation Logic in Java Solved the Power of Two problem on LeetCode! The challenge is to determine whether a given integer n is a power of two. The key insight lies in understanding the binary representation of numbers that are powers of two — they contain exactly one set bit (1). So, when we perform the operation n & (n - 1), it turns off the only set bit of n. For any power of two, this result becomes 0. Therefore, the condition n > 0 && (n & (n - 1)) == 0 accurately checks if a number is a power of two. This simple yet powerful bitwise trick is both efficient and elegant — a great reminder of how understanding binary operations can make problem-solving in programming much faster! ⚡ #Java #LeetCode #ProblemSolving #BitManipulation #CodingJourney #LearningEveryday

🚀 Accessing Static Members Using the Class Name in Java (Oop Concepts) In Java, static members are associated with the class itself, not with instances of the class. Consequently, you should always access static members using the class name, rather than an instance of the class or the `this` keyword. Using the class name clarifies that you are accessing a shared, class-level property or method. This promotes code clarity and avoids confusion about the scope of the variable or method. Learn more on our app: https://lnkd.in/gefySfsc #oopconcepts #programming #coding #tech #learning #professional #career #development

💡 𝗝𝗮𝘃𝗮/𝐒𝐩𝐫𝐢𝐧𝐠 𝐁𝐨𝐨𝐭 𝗧𝗶𝗽 - 𝗦𝘄𝗶𝘁𝗰𝗵 𝗘𝘅𝗽𝗿𝗲𝘀𝘀𝗶𝗼𝗻 💎 🕯 𝗧𝗿𝗮𝗱𝗶𝘁𝗶𝗼𝗻𝗮𝗹 𝗦𝘄𝗶𝘁𝗰𝗵 𝗦𝘁𝗮𝘁𝗲𝗺𝗲𝗻𝘁 The traditional switch statement has been part of Java since the beginning. It requires explicit break statements to prevent fall-through, which can lead to bugs if forgotten. Each case must contain statements that execute sequentially, making the code verbose and error-prone. 💡 𝗠𝗼𝗱𝗲𝗿𝗻 𝗦𝘄𝗶𝘁𝗰𝗵 𝗘𝘅𝗽𝗿𝗲𝘀𝘀𝗶𝗼𝗻 Switch expressions were introduced in Java 14 as a more concise and safe alternative. Using the -> syntax, you eliminate the need for break statements and can directly return values. Multiple cases can be grouped with commas, and the compiler enforces exhaustiveness for better safety. ✅ 𝗞𝗲𝘆 𝗕𝗲𝗻𝗲𝗳𝗶𝘁𝘀 ◾ No break statements, safer and cleaner code. ◾ Direct value assignment, treat switch as an expression. ◾ Multiple labels with comma separation. ◾ Compiler exhaustiveness checks, fewer runtime errors. 🤔 Which one do you prefer? #java #springboot #programming #softwareengineering #softwaredevelopment

Day 43 of #50DaysOfLeetCodeChallenge Problem: Valid Parentheses Approach: Used a stack to keep track of opening brackets. For each closing bracket, checked if it matches the top of the stack. Returned false if there was a mismatch or stack was empty. This ensures all brackets are properly nested and closed. Key Takeaways: Stacks are perfect for problems involving nested structures. Simple traversal with O(n) time complexity and O(n) space. Feeling more confident handling pointers and references in Java! Performance: Using a stack this way really helped me understand how to manage nested elements efficiently. #LeetCode #Java #DSA #CodingChallenge #ProblemSolving #Stack #Programming

Day 7 of #50DaysOfCode – Java Today I practiced arrays and conditional statements by writing a program to find the smallest number from five user inputs. 💻 What I did: Took 5 numbers from the user Stored them in an array Used a loop and if condition to find the smallest number 🧠 Concepts learned: Arrays (int[]) Loops (for) Conditional statements (if) Input/output using Scanner Step by step, getting stronger with Java! 💪 #Java #CodingJourney #LearnInPublic #Programming #50DaysOfCode

Did you know Java has its own mini garbage collector per thread? Not exactly, but it can feel that way. Each thread in Java has its own memory area (stack and local objects), while the JVM’s garbage collector manages cleanup concurrently across threads. That’s why one background thread may finish quickly, while another keeps the GC busy a little longer. Understanding how Java’s memory and GC threads interact can make you significantly better at debugging performance issues, especially when things behave unpredictably under load. Remember: “ Garbage collection isn’t magic , it’s just smart housekeeping. ” #Java #Programming #SoftwareEngineering #BackendDevelopment #Performance #JVM #MemoryManagement #LearnInPublic #DidYouKnowTech

Recently, I got curi🤔us about how high-level languages like Go, Java, or Rust are actually built. Do they all start from assembly? Turns out, not all high-level languages start from assembly. Take Go (Golang) as an example, the first Go compiler that turned Go code into machine code was written in C, not assembly. As Go matured, this C-based compiler was replaced by a new compiler written in Go itself. This fascinating process is called bootstrapping, it’s when a programming language’s compiler is written in the same language it compiles. In other words, the language learns to build itself. #golang #rustlang #bootstrapping #compiler #java #SoftwareDevelopment #Programming #GoDeveloper #SoftwareEngineering

Day 21 of #50DaysOfCode – Java Today’s challenge was Printing All Prime Numbers Between 1 and N. This task helped me understand how to identify prime numbers and how nested loops work in Java. Key learnings from today: 🔹 A prime number is divisible only by 1 and itself 🔹 We can check primality by testing divisibility 🔹 Using logical conditions and loops together improves problem-solving Slowly building confidence with new concepts every day! 💪✨ #Java #CodingJourney #50DaysOfCode #PrimeNumbers #LogicBuilding #ProgrammingBasics #LearningEveryday