How to Check String Prefix/Suffix in Java/Spring Boot

Day 47 of #100DaysOfLeetCode Today's problem: Reverse Words in a String Language used: Java What I did today: I implemented a solution to reverse the order of words in a given string while handling extra spaces efficiently. Used split("\\s+") to handle multiple spaces. Reconstructed the reversed string using StringBuilder. Applied trim() to clean up leading and trailing spaces. Key Takeaways: split("\\s+") is great for ignoring multiple spaces. Always use trim() to ensure a neat final output. StringBuilder makes string manipulation much faster than direct concatenation. #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #Programming #DeveloperJourney #LearnToCode

Hey Uplifters! 👋 Problem of the Day: (23-10-2025) 3461 - Check If Digits Are Equal in String After Operations I Problem in short: • You’re given a string s consisting of digits. • In each operation, replace every pair of consecutive digits with their sum mod 10, until only two digits remain. • Finally, return true if the last two digits are the same - otherwise false. 💡 Intuition: • We directly simulate the process as described - at every step, build a new string using the sum (mod 10) of adjacent digits until only two remain. • This is a beginner-friendly brute force approach that clearly shows the transformation process. 🧩 Approach: • Use a StringBuilder in java (Or normal string in cpp) for efficient string manipulation. • Repeatedly form a new string with (v1 + v2) % 10 for each adjacent pair. • Stop when only two digits are left, and compare them. Code Implementation with Explanation (Java): 👉 https://lnkd.in/g2rAty_T (GitHub) 👉 https://lnkd.in/gaygqRPt (Leetcode) Drop your thoughts below Let’s keep learning and growing together! 🌱 #LeetCode #POTD #ProblemOfTheDay #DailyChallenge #CodingChallenge #DataStructures #Algorithms #BinarySearch #ProblemSolving #Java #JavaProgramming #LearnToCode #CodeNewbie #TechCommunity #SoftwareDevelopment #Programming

💡 𝗝𝗮𝘃𝗮/𝐒𝐩𝐫𝐢𝐧𝐠 𝐁𝐨𝐨𝐭 𝐏𝐞𝐫𝐟𝐨𝐫𝐦𝐚𝐧𝐜𝐞 𝐓𝐢𝐩 🔥 💎 𝗣𝗿𝗲𝗳𝗲𝗿 𝗦𝘁𝗿𝗶𝗻𝗴𝗕𝘂𝗶𝗹𝗱𝗲𝗿 𝗢𝘃𝗲𝗿 𝗦𝘁𝗿𝗶𝗻𝗴 𝗖𝗼𝗻𝗰𝗮𝘁𝗲𝗻𝗮𝘁𝗶𝗼𝗻 𝗶𝗻 𝗟𝗼𝗼𝗽𝘀 🐌 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝘄𝗶𝘁𝗵 𝗦𝘁𝗿𝗶𝗻𝗴 𝗖𝗼𝗻𝗰𝗮𝘁𝗲𝗻𝗮𝘁𝗶𝗼𝗻  Strings are immutable in Java, so the '+' operator creates a new String object with every concatenation. In loops, this creates thousands of temporary objects that need to be garbage collected. This dramatically impacts both performance and memory usage. 🔥 𝗪𝗵𝘆 𝗦𝘁𝗿𝗶𝗻𝗴𝗕𝘂𝗶𝗹𝗱𝗲𝗿 𝗶𝘀 𝗕𝗲𝘁𝘁𝗲𝗿 StringBuilder uses a mutable character buffer and modifies it in place without creating new objects. In benchmarks with 10,000 iterations, StringBuilder completes in ~4ms while '+' operator takes ~400ms. That's 100x faster with significantly lower memory allocation. ✅ 𝗪𝗵𝗲𝗻 𝘁𝗼 𝗨𝘀𝗲 𝗘𝗮𝗰𝗵 ◾ Use StringBuilder for loops and multiple concatenations. ◾ Use '+' for simple, single-line string building (2-3 strings). ◾ The compiler optimizes simple '+' usage, but not in loops. #java #springboot #programming #softwareengineering #softwaredevelopment

🚀 Just solved LeetCode #2744 — Find Maximum Number of String Pairs 📘 Problem: You’re given an array of distinct strings. We need to count how many pairs (i, j) can be formed such that the string at index i is the reverse of the string at index j. Each string can only belong to one pair. Example: Input → ["cd", "ac", "dc", "ca", "zz"] Output → 2 Explanation → ✅ "cd" ↔ "dc" ✅ "ac" ↔ "ca" 🧠 My Approach: I used a HashSet to efficiently track reversed strings. 1️⃣ For each word, I created its reversed version. 2️⃣ If the reversed version already exists in the set, I counted a valid pair and removed it from the set. 3️⃣ Otherwise, I added the word to the set for future checks. 💡 What I Learned: ✅ How HashSet helps reduce time complexity for lookup and removal ✅ How to reverse strings efficiently using StringBuilder ✅ Strengthened logic-building for string manipulation and pairing problems 💻 Language Used: Java 🔍 Concept: HashSet | String Manipulation | Two-Pointer Logic #LeetCode #Java #DSA #String #HashSet #CodingUpdate #LearningByDoing #Programming

💻 LeetCode Challenge – Day 5: String to Integer (atoi) Today's problem was all about converting a string into an integer — seems simple, but the edge cases make it tricky! 🧠 🔹 Problem: Implement the atoi function, which converts a string to a 32-bit signed integer (just like in C/C++). 🔹 Key Learnings: ✅ Handling whitespaces and optional '+' or '-' signs ✅ Managing non-digit characters gracefully ✅ Preventing integer overflow & underflow ✅ Importance of clean parsing logic This problem really improved my understanding of string manipulation and boundary conditions in Java. 🚀 #LeetCode #100DaysOfCode #Java #CodingChallenge #ProblemSolving #StringToInteger #CodingJourney

💻 Did you know you can print output in Java without a main() method? Here’s how 👇 class WithoutMain { static int variable = 10; static { System.out.println(variable); System.exit(0); } } 🧠 Explanation: • When the JVM loads the class, all static blocks execute automatically — even before checking for the main() method. • The System.exit(0); line stops execution before the JVM looks for main(), preventing the usual “main method not found” error. ✅ Output: 10 This is a fun trick to understand how class loading and static blocks work in Java. Not practical for production — but a great interview or concept question! 😉 #Java #Programming #LearnWithHarsh #CodingFacts #JVM #TechLearning

📌 Day 33/100 – Check If Digits Are Equal in String After Operations (LeetCode 3461) 🔹 Problem: Given a string s consisting of digits, repeatedly replace it by a new string formed by taking the sum (mod 10) of every pair of adjacent digits until only two digits remain. Return true if the final two digits are the same, otherwise false. 🔹 Approach: Iteratively built a new string using adjacent digit sums modulo 10. Repeated the process until the string length reduced to 2, then compared the final digits. Used StringBuilder for efficient construction of intermediate strings. 🔹 Key Learning: Reinforced clean looping and string handling patterns in Java. 🔹 Complexity: Time: O(n²) — Each iteration processes a smaller string. Space: O(n) for intermediate string storage. #100DaysOfCode #LeetCode #Java #StringManipulation #DSA #ProblemSolving #CodingPatterns #Iteration #ModularArithmetic

Ever debugged for hours only to realize you were comparing Strings with == instead of .equals()?  You're not alone. 🤦♂️ The culprit? Java's String Pool - a memory optimization that trips up even experienced developers. I just wrote a deep dive into: ✅ How String Pool actually works behind the scenes ✅ Why comparing the same content with == sometimes returns true, sometimes false ✅ The intern() method and when to use it ✅ Real performance benchmarks ✅ Common pitfalls that waste memory Practical code examples included. 5-minute read. Read here: https://lnkd.in/g4emECJM #Java #Coding #Programming #SoftwareDevelopment #SoftwareEngineering

🚀 Java Hack: Jagged Arrays Made Simple! Did you know 2D arrays in Java don’t have to be rectangular? 🤯 Each row can have a different number of elements – that’s called a jagged ✨ Why Jagged Arrays? 1.Flexible row sizes 2.Memory efficient when rows vary 3.Perfect for irregular data structures #Java #CodingTips #Programming #LearnJava #JaggedArray #TechInsightss