Choosing the right Java collection for your needs: List, Set, LinkedHashSet.

#Day-44) of Problem Solving | LeetCode 2011 – Final Value of Variable After Performing Operations Just solved a fun one! 🚀 This problem tests how well you can interpret simple string-based operations and apply them efficiently. Given a list of operations like "++x", "x--", etc., the goal is to compute the final value of a variable starting from zero. 🔍 My approach (Java): Used String.indexOf("+") to detect increment operations and updated the counter accordingly. Clean, readable, and runs in linear time. #LeetCode #Java #ProblemSolving #DSA #CodingChallenge #PranshuCodes #LinkedInCoding #TechJourney

🚀 Day 11 of 100 Days of LeetCode! 💻 Today’s problem: Implement strStr() 🔍 🧩 Problem #28: Find the Index of the First Occurrence in a String This problem is a classic example of string pattern matching, where we need to find the starting index of a substring (needle) in a given string (haystack). ✨ My Approach: Used a simple sliding window technique to check each substring of haystack with length equal to needle. Compared it directly using equals() to find a match. Time complexity: O((n - m + 1) * m), which is acceptable for moderate input sizes. ✅ Result: All test cases passed successfully — and achieved 100% runtime efficiency ⚡ Each day I’m learning to think more efficiently and write cleaner, more optimized code. Consistency really is the key 🔑 #Day11 #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingJourney #SoftwareDevelopment #DSA

🚀 Day 59 of #100DaysOfCode 🚀 🔹 Problem: Check if Digits Are Equal in String After Operations I – LeetCode ✨ Approach: Used an iterative reduction strategy 🔁 — repeatedly combined adjacent digits (mod 10) until only two numbers remained. Finally checked if both digits are equal! Simple yet logical 🧠 ⚡ Complexity Analysis: Time Complexity: O(n²) – iterative pairwise reduction until only two digits remain Space Complexity: O(n) – storing intermediate list of digits 📊 Performance: ✅ Runtime: 10 ms (Beats 35.69%) ✅ Memory: 45.51 MB (Beats 12.86%) 🔑 Key Insight: Sometimes, brute-force reduction problems aren’t about optimization — they’re about translating logic into clean code that mirrors the operation flow perfectly. ✨ #LeetCode #100DaysOfCode #Java #DSA #ProblemSolving #CodingChallenge #LogicBuilding #ProgrammingJourney #DailyCoding

💻 Day 410 of #500DaysOfCode 🚀 Problem: Minimum One Bit Operations to Make Integers Zero (LeetCode #1611) Difficulty: Hard Today’s challenge was all about bit manipulation — one of those topics that looks simple on the surface but reveals some deep patterns when you dig in. The task: Given an integer n, you have to transform it into 0 using specific bit operations — flipping bits based on certain conditions. At first glance, it seems like a direct simulation problem, but the optimal approach lies in understanding Gray Code transformations! 💡 Key Idea: The number of operations needed to reduce n to 0 follows an inverse Gray code pattern. Using recursion and bitwise manipulation, we can calculate the answer efficiently in O(log n) time. ✨ Concepts Learned: Gray code transformation patterns Recursive bit manipulation How mathematical patterns simplify complex bit problems Each hard problem strengthens logical thinking a bit more 🔥 #Day410 #LeetCode #BitManipulation #Java #CodingChallenge #ProblemSolving #CodeEveryday #500DaysOfCode

#Day 8 of the LeetCode Challenge: Crushing "Power of Two" (Problem 231) with a Bitwise Bang! Today marks Day 8 of my commitment to consistent LeetCode practice, and I tackled an "easy" problem that reveals a deep secret about integers: #Power of Two (Problem 231) of LeetCode. While there are multiple ways to solve this, I went straight for the most efficient technique: Bitwise Manipulation. The trick lies in the expression: n & (n - 1). The Power of Bitwise (&) Any positive power of two (like 8 or 16) has only one bit set to '1' in its binary representation. Subtracting 1 from it flips that single '1' to '0' and all trailing '0's to '1's. When you perform a bitwise AND (&) between n and n-1, the result is {0} if, and only if, n was a power of two. The result? An elegant, O(1) time complexity solution that's often faster than looping or using logarithms! Code Snippet (Java): Java public boolean isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; } Seeing that 1ms runtime on the platform makes the small investment in learning bitwise operators totally worth it. These small optimizations are what separate good code from great code. What's a clever bitwise trick you rely on? Share your favorite optimization in the comments! 👇 #LeetCode #CodingChallenge #ProblemSolving #Java #BitwiseOperators #TechSkills #Day8 #SoftwareEngineering

🎯 Day 8 of #365DaysofDSA Today I solved “Convert Sorted Array to Binary Search Tree” (LeetCode #108) 🌲 💡 Concepts used: • Recursion • Divide and Conquer • Binary Search Tree Construction ✨ Approach Summary: The goal is to convert a sorted array into a height-balanced BST. To maintain balance, I picked the middle element of the array as the root — ensuring roughly equal elements on both sides. Then recursively repeated the process for left and right halves of the array. For each recursive call: 1️⃣ Find the middle index of the current subarray. 2️⃣ Create a node with that middle element. 3️⃣ Recursively build left and right subtrees using the respective halves. 🚀 Key takeaway: The divide and conquer strategy naturally fits problems that require balance and recursion — just like in this BST construction. Breaking a big problem into symmetric subproblems leads to elegant and efficient solutions. 🌿 #Day8 #DSA #LeetCode #Java #BinarySearchTree #Recursion #ProblemSolving #CodingJourney #LearnByDoing #Programming

💻 Day 59 of #100DaysOfCode Challenge 🚀 Today's problem: Leetcode 240 – Search a 2D Matrix II 🔍 This problem was about efficiently searching for a target value in a 2D matrix where each row and each column is sorted in ascending order. 💡 Key Insight: Instead of searching the entire matrix, start from the top-right corner: If the current element is greater than the target → move left. If it’s smaller → move down. This approach ensures that we eliminate one row or one column in every step — achieving O(m + n) time complexity. 📊 Result: ✅ Accepted – All test cases passed successfully! ⚡ Time Complexity: O(m + n) 💾 Space Complexity: O(1) 🧠 Takeaway: This problem was a great reminder that understanding data structure properties can help you find smarter solutions rather than brute-forcing through the problem. Every problem adds another brick to the foundation of logical thinking and algorithmic mastery. 💪 #Day59 #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #TechLearning #DSA

🚀 Day 08/10 of Linked List Series: Delete a Node by its Index🎯 🧠 Algorithm (Step-by-Step Explanation) 1️⃣ Start from the head of the linked list. 2️⃣ If the index = 0, it means we want to delete the first node — so just move the head to the next node. 3️⃣ Otherwise, use a loop to reach the node just before the target index (that is, index - 1). 4️⃣ Once you reach there, skip the next node by connecting the current node’s next pointer to the node after the one being deleted. 5️⃣ Return the (possibly updated) head of the linked list. 🪄 Simple Understanding: Imagine a chain of people holding hands (each one pointing to the next). If you want to remove the person at position index, just tell the person before them to hold the hand of the one after them — the middle one is now “unlinked” and effectively deleted! 🔥 Keep Learning — Keep Building #CodeWithLakkojuEswaraSai #CodeWithLakkojuEswaraSai_LinkedList #Java #DSA #DataStructures #LinkedList #CodingCommunity #100DaysOfCode #ProgrammerLife #LearnWithMe #CodeEveryday #TechJourney #CodingIsFun #JavaDevelopers #SoftwareEngineering #CodeNewbie

🚀 Day 69 of My LeetCode Journey 🚀 Problem: Remove Linked List Elements Today’s problem reinforced one of the most important concepts in Linked Lists — safe node deletion using pointers. 💡 What I learned today: Sometimes the approach looks right… …until you realize a tiny mistake in logic is breaking the whole solution. Fixing that one small piece can completely turn the code around. 🧠 Key takeaway: Don’t rush to write code. Sit back → think → visualize → break the problem → then solve. This simple switch saves half the debugging time. #LeetCode #Day69 #CodingJourney #Java #DSA #ProblemSolving #Consistency

📌 Day 4/100 - Minimum Size Subarray Sum (LeetCode 209) 🔹 Problem: Given an array of positive integers and a target value, find the minimal length of a contiguous subarray whose sum is greater than or equal to the target. If there’s no such subarray, return 0. 🔹 Approach: Used the Sliding Window technique for an optimized solution: Initialize two pointers (low, high) and a running sum. Expand the window by moving high until the sum ≥ target. Once valid, shrink the window from the left to find the smallest subarray. Keep updating the minimum length throughout. This reduced the time complexity from O(n²) (brute force) to O(n). 🔹 Key Learning: Sliding Window is ideal for problems with contiguous subarrays. Optimization often comes from adjusting the window efficiently. Each problem strengthens logical flow and pattern recognition. Another step forward in mastering DSA and problem-solving consistency! ⚡ #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingChallenge #SlidingWindow