Converting Roman Numerals to Integers in Java

🚀 Day 72 — LeetCode Practice 🚀 Today’s problem: Sorting the Sentence 💡 What I learned today: • Practiced string manipulation and splitting techniques • Understood how to extract numbers from characters using ASCII logic • Learned to map words to correct positions using indexing • Improved attention to detail while handling edge cases 🧠 Key idea: Each word has a number at the end → use it to place the word in the correct position → then remove the number and rebuild the sentence. ✨ Consistency is slowly turning confusion into clarity. #Day72 #LeetCode #Java #ProblemSolving #DSA #CodingJourney #Consistency

🚀 Day 7 of #LeetCode Challenge 🔍 Problem: Sqrt(x) 💡 Approach: Used the built-in Math.sqrt() function to compute the square root and then typecasted it to an integer to get the floor value. 🧠 Key Concept: Understanding how to convert a floating-point result into an integer using typecasting. 🔥 #Day7 #LeetCode #Java #DSA #CodingJourney #Consistency

Solved: Linked List Cycle (LeetCode 141) Today’s problem was all about detecting cycles in a linked list using the classic Floyd’s Cycle Detection Algorithm (Tortoise & Hare) 🐢🐇 💡 Key Takeaways: Using two pointers (slow & fast) makes cycle detection efficient No extra space needed → O(1) space complexity Time complexity stays optimal → O(n) Simple logic, but powerful concept for interviews Consistency is starting to pay off. Small steps every day = big progress over time. #Day25 #DSA #LeetCode #CodingJourney #Java #100DaysOfCode #ProblemSolving #TechGrowth

🚀 Day 3 of #LeetCodeDailyChallenge Showing up daily, even when the problem pushes me out of my comfort zone. Today’s problem: Longest Substring Without Repeating Characters 📌 Topic: Sliding Window, HashMap ⚡ Difficulty: Medium What I learned today: How the sliding window technique optimizes substring problems Using a HashMap to track last seen positions of characters Adjusting the window smartly to avoid rechecking elements Thinking in terms of O(n) optimization instead of brute force This one really changed how I approach string problems — from checking everything to working smarter with patterns. #DSA #LeetCode #Java #SlidingWindow #ProblemSolving #LearningInPublic #Consistency

Day 90 of #100DaysOfCode – Unique Binary Search Trees II Today’s problem was all about generating all structurally unique BSTs for values from 1 to n. At first glance, it feels tricky because it's not just counting trees — we actually need to build every possible tree structure. 💡 Key Insight: Pick each number i as the root Recursively build: Left subtree from [1 ... i-1] Right subtree from [i+1 ... n] Combine every left & right subtree pair 🌳 This is a classic Recursion + Backtracking problem and is closely related to Catalan Numbers. 📌 Example: For n = 3, we get 5 unique BSTs ⚡ What I learned today: How recursion can generate combinations of structures Importance of base case (start > end → null) Combining subproblems effectively 💻 Time Complexity: Approximately O(4ⁿ / √n) Consistency is key — 90 days strong and still going 💪 Next target: 💯 #DSA #Java #LeetCode #Recursion #BinaryTree #CodingJourney #100DaysOfCode

🚀 Day 33 of #128DaysOfCode 🧩 Problem Insight: The goal was to check whether a given string can be formed by repeating one of its substrings multiple times. 💡 Key Learning: Instead of checking all possible substrings (which can be inefficient), I learned an elegant trick: By concatenating the string with itself and removing the first and last characters, we can determine if the original string exists within it. ⚡ This approach helped me: - Improve my understanding of string patterns - Learn a smart optimization technique - Avoid brute-force solutions 🛠️ Concepts Practiced: - String manipulation - Pattern recognition - Optimized problem-solving approach 📈 Every day I’m getting better at identifying patterns and writing cleaner, more efficient code. #Day33 #128DaysOfCode #Java #DSA #CodingJourney #ProblemSolving #LeetCode

🚀 Day 26/100 Days of #CodeChallenge Today’s problem: Isomorphic Strings (Leetcode 205) This problem helped me understand how to map characters between two strings while maintaining consistency and order. 💡 Key Concept: Two strings are isomorphic if characters in one string can be replaced to get the other string — with: ✔️ One-to-one mapping ✔️ No two characters mapping to the same character ✔️ Order preserved 🧠 What I Learned: How to use mapping (arrays/hashmaps) efficiently Importance of bidirectional checking Handling edge cases like unequal lengths ⚡ Approach: Compare lengths first Track character mappings using arrays Ensure consistency in both directions ⏱️ Complexity Analysis: Time Complexity: O(n) → We traverse the strings once Space Complexity: O(1) → Fixed-size arrays (256 characters) ✅ Successfully solved and understood the logic! Every day is a step closer to mastering problem-solving 💪 #Day26 #100DaysOfCode #Java #DSA #LeetCode #CodingJourney #ProblemSolving #TechLearning

Day 35 of Consistency 💻🔥 Solved Longest Substring Without Repeating Characters — a classic sliding window problem that really tests optimization skills. ✨ Key Takeaways: Mastered the sliding window technique Used HashMap for efficient lookup Improved understanding of two-pointer approach ⚡ Performance: Runtime: 5 ms Beat: 87%+ submissions Every day is about getting sharper, not just solving problems. #LeetCode #DataStructures #Algorithms #CodingJourney #Consistency #Java #ProblemSolving

#Day360 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Search in a Binary Search Tree (LeetCode 700) 💡 Approach: Used the properties of a Binary Search Tree to efficiently search for a value. At each step: If the value is smaller, move to the left subtree If larger, move to the right subtree This reduces the search space at every step. ⏱ Time Complexity: O(h) 🧠 Space Complexity: O(h) Continuing daily consistency in problem solving 🚀 #DSA #Java #LeetCode #BinaryTree #ProblemSolving #Coding

Day 72 of #100DaysOfCode Problem: Convert Sorted Array to Height-Balanced BST Today I learned how to efficiently convert a sorted array into a balanced Binary Search Tree using Divide & Conquer. Key Insight: Pick the middle element as the root to maintain balance. Recursively build: Left subtree from left half Right subtree from right half This ensures: Optimal height Faster search operations ⏱ Time Complexity: O(n) 📦 Space Complexity: O(log n) Consistency is the real game changer #DSA #Java #BinaryTree #LeetCode #CodingJourney