Convert Integer to Roman Numeral

Leetcode Practice - 15. 3Sum The problem is solved using JAVA. Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.   Example 1: Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter. Example 2: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0. Example 3: Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.   Constraints: 3 <= nums.length <= 3000 -105 <= nums[i] <= 105 #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

Day 7 of Java with DSA Journey 🚀 📌 Topic: Missing Number (LeetCode 268) Quote: "Sometimes the best way to solve a coding problem is to put down the keyboard and pick up a calculator." ✨ What I learned today: 🔹 Mathematical Optimization: Instead of brute force, I used a simple formula to compute the expected sum. {Sum} = \frac{n(n+1)}{2} 🔹 Core Idea: Expected Sum − Actual Sum = Missing Number 🔹 Alternative Approach: Used XOR technique where duplicates cancel out → leaving the missing value. 🔹 Efficiency: ✔️ Time Complexity: O(n) ✔️ Space Complexity: O(1) 🧠 Problem Solved: ✔️ Missing Number 💡 Key Insight: We often jump to HashSet or Sorting, but this problem taught me to pause and think mathematically first. A simple formula reduced space complexity from O(n) → O(1) 🚀 ⚡ Interview Tip (Important!): Be careful of integer overflow in Java: n * (n + 1) can exceed int limit Use long OR switch to XOR method for safety Consistency is the real key 🔑 #DSA #LeetCode #Java #Mathematics #CodingJourney #ProblemSolving #Day7 #Algorithms #Optimization #Array #MCA #lnct #100DaysOfCode #Algorithms #SoftwareEngineering #InPlaceAlgorithms #TechLearning #JavaDeveloper

Leetcode Practice - 16. 3Sum Closest The problem is solved using JAVA Given an integer array nums of length n and an integer target, find three integers at distinct indices in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.   Example 1: Input: nums = [-1,2,1,-4], target = 1 Output: 2 Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). Example 2: Input: nums = [0,0,0], target = 1 Output: 0 Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).   Constraints: 3 <= nums.length <= 500 -1000 <= nums[i] <= 1000 -104 <= target <= 104 #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Postorder Traversal (LeetCode 145) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform postorder traversal of the binary tree. • Recursively traverse the left subtree • Recursively traverse the right subtree • Visit the root node and store its value • Follow Postorder pattern: Left → Right → Root This approach works efficiently because recursion naturally follows the postorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the postorder pattern (Left → Right → Root), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Inorder Traversal (LeetCode 94) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform inorder traversal of the binary tree. • Recursively traverse the left subtree • Visit the root node and store its value • Recursively traverse the right subtree • Follow Inorder pattern: Left → Root → Right This approach works efficiently because recursion naturally follows the inorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the inorder pattern (Left → Root → Right), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal

🔥 𝗗𝗮𝘆 𝟵𝟰/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟱𝟯𝟵. 𝗞𝘁𝗵 𝗠𝗶𝘀𝘀𝗶𝗻𝗴 𝗣𝗼𝘀𝗶𝘁𝗶𝘃𝗲 𝗡𝘂𝗺𝗯𝗲𝗿 | 🟢 Easy | Java Marked as Easy — but the optimal solution is pure binary search brilliance. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Given a sorted array, find the kth missing positive integer. Linear scan works — but can we do O(log n)? 💡 𝗧𝗵𝗲 𝗞𝗲𝘆 𝗜𝗻𝘀𝗶𝗴𝗵𝘁 At index i, the value arr[i] should be i+1 in a complete sequence. So missing numbers before arr[i] = arr[i] - 1 - i This lets us binary search on the count of missing numbers! ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 ✅ If arr[mid] - 1 - mid < k → not enough missing numbers yet, go right ✅ Else → too many missing, go left ✅ After the loop, left + k gives the exact answer 𝗪𝗵𝘆 𝗹𝗲𝗳𝘁 + 𝗸? After binary search, left is the index where the kth missing number falls beyond. left numbers exist in the array before that point, so the answer is left + k. No extra passes needed. ✨ 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(log n) — vs O(n) linear scan 📦 Space: O(1) This is a perfect example of binary searching on a derived condition, not just a value. A real upgrade from the naive approach. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gVYcjNS6 𝟲 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗧𝗵𝗲 𝗳𝗶𝗻𝗶𝘀𝗵 𝗹𝗶𝗻𝗲 𝗶𝘀 𝗿𝗶𝗴𝗵𝘁 𝘁𝗵𝗲𝗿𝗲! 🏁 #LeetCode #Day94of100 #100DaysOfCode #Java #DSA #BinarySearch #Arrays #CodingChallenge #Programming

Day 4 of my DSA Journey 🚀 Today’s challenge: The classic Palindrome Number problem. For today's solution, I focused on an intuitive approach using String manipulation in Java. Sometimes the best way to solve a problem is to break it down into steps that make logical sense right out of the gate! 🧠 My Approach: Handle the edge case: Negative numbers can never be palindromes, so return false immediately. Convert the integer to a String. Use a for loop to iterate backwards through the string, building a new reversed string character by character. Compare the original string with the reversed string to determine if it's a palindrome. ⚡ Key Learnings & Java Gotchas: String Iteration: In Java, strings aren't exactly like arrays, so I had to use .charAt(i) instead of standard bracket notation [i] to grab individual characters. The Comparison Trap: I learned the crucial difference between == and .equals(). In Java, == compares memory locations, but .equals() checks if the actual text values are the same. A huge "aha!" moment for handling Strings! Edge Case Handling: Always check for negative numbers first to save processing time. 📌 Complexity: Time: O(n) — where n is the number of digits, since we loop through the string once. Space: O(n) — for storing the string representations of the number. Every bug is a lesson, and every solved problem is a step forward. On to the next one! 💪 #DSA #DataStructures #Algorithms #ProblemSolving #CodingJourney #Java #TechJourney#DSAJourney #LeetCode #Coding #LearnInPublic #GrowthMindset

Solved Find Peak Element -> LeetCode(162) Medium, Binary Search in Java. Till now I had solved around 10-11 binary search problems. But in all of them array was always sorted , even rotated ones had at least one sorted half. So I had one strong assumption , binary search only works on sorted arrays. This problem broke that assumption completely. No sorted half, no rotation logic working. I was stuck because none of my previous patterns were fitting here. Took help from Claude. It suggested slope based thinking , if right neighbour is greater, peak is on right side, go right. If left neighbour is greater, go left. If both neighbours are smaller, current element is peak. Then I questioned > what if slope breaks in between? Claude pointed me to re-read the problem. The key insight was that array has -∞ at both ends virtually, which guarantees at least one peak always exists. Applied my own logic from there and got it accepted Then Claude showed me a cleaner version , when low < high, at the point where low == high we already have our peak. No need for extra boundary checks. That gave me a second shorter solution. Also broke my second assumption > binary search doesn't always need while(low <= high). Sometimes while(low < high) is cleaner. Two wrong assumptions fixed in one problem 🙌 Took help but questioned it, understood it, then coded it myself. Github Repo link : https://lnkd.in/grF5ACw5 **Any other wrong assumption you had about binary search? Drop it below 👇** #DSA #Java #LeetCode #BinarySearch #LearningInPublic

𝗠𝗮𝘀𝘁𝗲𝗿𝗶𝗻𝗴 𝗝𝗮𝘃𝗮’𝘀 𝘃𝗮𝗿 Recently, I explored how Java introduced Local Variable Type Inference (var) in Java 10 and how it transformed the way developers write cleaner and more expressive code. Java has traditionally been known for its verbosity. With the introduction of var through Project Amber, the language took a major step toward modern programming practices—balancing conciseness with strong static typing. 𝗪𝗵𝗮𝘁 𝗜 𝗹𝗲𝗮𝗿𝗻𝗲𝗱: • var allows the compiler to infer types from initializers, reducing boilerplate without losing type safety. • It is not a keyword, but a reserved type name, ensuring backward compatibility. • Works only for local variables, not for fields, method parameters, or return types. • The inferred type is always static and compile-time resolved—no runtime overhead. • Powerful in handling non-denotable types, including anonymous classes and intersection types. Must be used carefully: • Avoid when the type is unclear from the initializer • Prefer when the initializer clearly reveals the type (e.g., constructors or factory methods) • Enhances readability only when the initializer clearly conveys the type. 𝗞𝗲𝘆 𝘁𝗮𝗸𝗲𝗮𝘄𝗮𝘆: var is not just about writing less code—it’s about writing clearer, more maintainable code when used correctly. The real skill lies in knowing when to use it and when not to. A special thanks to Syed Zabi Ulla sir at PW Institute of Innovation for their clear explanations and continuous guidance throughout this topic. #Java #Programming #SoftwareDevelopment #CleanCode #Java10 #Developers #LearningJourney