"Removing Duplicates from Sorted Linked List in Java"

🚀Day 96/100 #100DaysOfLeetCode 🧩Problem: Reverse Linked List II✅ 💻Language: Java 💡 Approach: 1️⃣ First, use a dummy node to handle edge cases where reversal starts at the head. 2️⃣ Traverse to the node just before the left position — call it prev. 3️⃣ Reverse the sublist between left and right using standard pointer manipulation. 4️⃣ Reconnect the reversed portion back into the original list. 🔑Key Takeaways: 🔹Dummy nodes simplify linked list edge cases. 🔹In-place reversal reduces memory overhead. 🔹Careful pointer tracking ensures list integrity. ⚙️Performance: ⏱️Runtime: 0 ms(beats 100.00%) 💾Memory: 41.29 MB(beats 70.37%) #100DaysOfLeetCode #Java #LinkedList #CodingJourney #ProblemSolving #DSA #LeetCode #CodingChallenge

Day 91 of #100DaysOfCode Solved Base 7 in Java 🔢 Approach The challenge was to convert a given integer (num) to its base 7 string representation. Conversion Method The core of the solution lies in the standard algorithm for base conversion: repeated division and remainder collection. Handle Zero and Negatives: If the input num is 0, the result is immediately "0". I determined if the number is negative and stored this in a boolean flag, then proceeded with the absolute value of the number (num = Math.abs(num)) for the conversion logic. Conversion Loop: I used a while loop that continues as long as num > 0. In each iteration: The remainder when num is divided by 7 (num % 7) gives the next digit in base 7. This digit is appended to a StringBuilder. num is then updated by integer division by 7 (num /= 7). Final Result: Since the remainders are collected in reverse order, I called sb.reverse(). If the original number was negative, I prepended a hyphen (-) to the reversed string. Finally, I returned the result as a string. This simple and efficient implementation had a very fast runtime, beating 77.39% of submissions. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #BaseConversion #ProblemSolving

(Largest Perimeter Triangle — Java 🔺) Hey everyone 👋 Today I explored how to find the largest perimeter triangle from a given array of side lengths using Java. While solving this, I understood how a simple mathematical condition can help in optimizing the entire logic 🔥 Here’s the breakdown 👇 🧠 Concept: To form a valid triangle from 3 sides, the sum of any two sides must be greater than the third one. → a + b > c → a + c > b → b + c > a 💡 Approach 1: Sort the array in descending order and check triplets from the start. If any 3 consecutive sides satisfy the triangle condition, their sum gives the largest perimeter. ⚡ Approach 2 (Optimized): Instead of reversing the array, just sort it in ascending order and iterate from the end — gives the same result but with cleaner logic and less work. 🧩 Time Complexity: O(N log N) 💾 Space Complexity: O(1) This problem helped me understand how sorting order and traversal direction can make an algorithm more elegant and efficient 🚀 📁 Code is available on my GitHub repo: https://lnkd.in/ekjD9s22 #Java #DSA #ProblemSolving #Arrays #CodingJourney #Developers #LearningByDoing #FullStackDeveloper #MCA #GitHub

𝗝𝗮𝘃𝗮 𝗧𝗶𝗽: 𝗕𝘂𝘀𝘁𝗶𝗻𝗴 𝗮 𝗖𝗼𝗺𝗺𝗼𝗻 𝗠𝗲𝗺𝗼𝗿𝘆 𝗠𝘆𝘁𝗵 Let's talk about a common assumption in Java: if a short primitive is smaller than an int, then a Short object must be smaller than an Integer object, right? Not exactly. This is a great example of how Java's object model works. While short (2 bytes) is half the size of int (4 bytes), their wrapper classes tell a different story. Every object in Java comes with overhead—a header for metadata. This overhead is often larger than the data itself! On a standard 64-bit system, this header plus memory padding means that Byte, Short, and Integer objects all typically occupy the same 16 bytes of memory. So, what's the lesson? ✅ Use Integer by default: For collections like List<Integer>, it's the standard and doesn't waste memory compared to Short or Byte. ✅ Use Short or Byte for clear communication: They are great for showing that a field has a limited, known range. ✅ For real memory savings, use primitive arrays: A short[] is far more memory-efficient than an ArrayList<Short>. It's a small detail, but understanding it helps us write better, more intentional code. #Java #ProgrammingTips #SoftwareEngineering #CodeQuality #Developer

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🔢 Why Does 0123 Print as 83 in Java? 🤔 While working on constructors today, I came across an interesting behavior in Java that reminded me how subtle details in syntax can completely change what your code does! When I wrote this line 👇 Student objectTwo = new Student(0123); I expected it to print 123. But instead, the console output was: 83 So what’s happening here? 💡 In Java, when a number starts with a leading zero (0), it is interpreted as an octal (base 8) number — not a decimal one. Let’s decode it: 0123 (octal) = 1×8² + 2×8¹ + 3×8⁰ = 64 + 16 + 3 = 83 (decimal) Hence, Java prints 83! --- 🧩 Takeaway: ✅ 123 → Decimal (Base 10) ✅ 0123 → Octal (Base 8) ✅ 0x123 → Hexadecimal (Base 16) ✅ 0b1010 → Binary (Base 2) --- 💬 Lesson: Tiny syntax details can make a big difference. Always watch out for leading zeros in numeric literals — they might silently convert your values to something unexpected! --- 🔖 #Java #ProgrammingTips #Developers #CodeLearning #JavaBasics #CodingCommunity #SoftwareEngineering #TechLearning

#Day-68) LeetCode #3228 – Maximum Number of Operations to Move Ones to the End (Java Edition) Tackled this neat string problem using a greedy approach in Java. The challenge? Move '1's to the end of the string under specific movement rules, maximizing the number of valid operations. 🧠 Core Idea: Track how many '0's we've seen and how many '1's we've already moved. Only move a '1' if there's enough '0's to justify it and the next character is also '1'. 💻 Java Strategy: Loop through the string Use counters to manage '0's and '1's Let me know how you'd tweak this or if you see a more optimal path! #Java #LeetCode #GreedyAlgorithm #StringProblems #DSA #CodingChallenge #LinkedInTech #ProblemSolving

Today I explored, how sorting works in Java 8 using Lambdas and Streams. Here what I learned, List.of() creates an immutable list, so we can't modify it directly. If we want to do any modification on immutable list, we can 1. Create a mutable list List<Integer> immutableList = List.of(5, 8, 3, 2, 7, 1); List<Integer> mutableList = new ArrayList<Integer>(immutableList); mutableList.sort((a, b) -> a.compareTo(b)); 2. By using streams, can get new modified list. List<Integer> immutableList = List.of(5, 8, 3, 2, 7, 1); List<Integer> sortedList = immutableList.stream(). sorted(). collect (Collectors.toList); So, streams in Java 8 are non-mutating, they create a new result instead of modifying original data source. It feels great to understand the subtle differences between mutating (list.sort()) operations and functional operations (stream().sorted()). #Java #Java8 #Streams #Lambdas #CodingJourney #LearningEveryday #SoftwareDevelopment