Java LeetCode 482 License Key Formatting

Day 97 of #100DaysOfCode 💻 📌 LeetCode Problem 482 – License Key Formatting Today’s problem focused on string manipulation and formatting logic. 🔹 The task: Reformat a license key string so that: All letters are uppercase Groups are separated by - Each group has exactly k characters, except possibly the first 🔹 What I practiced: Traversing strings from the end Using StringBuilder efficiently Handling edge cases like trailing dashes and mixed characters 🔹 Approach: Remove all existing dashes Convert letters to uppercase Build new groups from the end and reverse the result 📚 Language: Java 🧠 Concepts: Strings, StringBuilder, Edge Cases, Formatting 97 days down — almost at the finish line! 💪🔥 #Day97 #LeetCode #Java #DSA #100DaysChallenge #StringManipulation #CodingJourney #Consistency

Day 97 of #100DaysOfCode 💻 📌 LeetCode Problem 482 – License Key Formatting Today’s problem focused on string manipulation and formatting logic. 🔹 The task:  Reformat a license key string so that: All letters are uppercase Groups are separated by - Each group has exactly k characters, except possibly the first 🔹 What I practiced: Traversing strings from the end Using StringBuilder efficiently Handling edge cases like trailing dashes and mixed characters 🔹 Approach: Remove all existing dashes Convert letters to uppercase Build new groups from the end and reverse the result 📚 Language: Java  🧠 Concepts: Strings, StringBuilder, Edge Cases, Formatting 97 days down — almost at the finish line! 💪🔥  #Day97 #LeetCode #Java #DSA #100DaysChallenge #StringManipulation #CodingJourney #Consistency

🚀 LeetCode #392 | Is Subsequence Solved LeetCode #392 – Is Subsequence using the Two Pointer technique in Java. 🔍 Core Idea : Traverse both strings simultaneously and try to match characters in order. 🧠 Logic Breakdown: Initialize two pointers sp → pointer for string s tp → pointer for string t Traverse string t from left to right If s.charAt(sp) == t.charAt(tp) → move sp forward Always move tp forward At the end, check whether all characters of s are matched ✅ If sp == s.length() → s is a subsequence of t ⏱ Complexity Time: O(n) Space: O(1) ✔ Accepted ⚡ Clean and efficient solution #LeetCode392 #LeetCode #TwoPointers #Java #DSA #ProblemSolving #CodingJourney #Consistency #FullStackDevelopment

🚀 Day 77 of LeetCode Challenge ✅ Problem: Add Strings 🔗 https://lnkd.in/gZatGJzA 💡 Approach: Traverse both strings from right to left Convert characters to digits and add them with a carry Append the result digit-by-digit using StringBuilder Reverse the final string to get the correct sum ⚡ Performance: ✅ All test cases passed ⏱ Runtime: 2 ms (Beats 91.47%) 💾 Memory efficient and clean implementation 📚 Key Learning: When direct numeric conversion isn’t allowed, manual digit-by-digit addition is a reliable and efficient strategy for handling large numbers stored as strings. 🔧 Language: Java 📈 Consistency over intensity—one problem at a time. #LeetCode #DSA #Java #StringManipulation #MathProblems #DailyCoding #Consistency #Day77

#Day_62 Problem: Roman to Integer (LeetCode 13) Converting Roman numerals into integers looks easy, but the subtraction rules make it interesting! 🏛️ 🧠 Insights: • Roman numerals usually add values left to right • Subtraction happens when a smaller value appears before a larger one • Traversing from right to left simplifies the logic ⚙️ Approach Used: • Store Roman symbols and values in a HashMap • Start from the last character • If current value < next value → subtract • Else → add to total 📌 What this improved: • HashMap usage • String traversal logic • Pattern recognition in number systems ✨ Understanding the rule is more important than memorizing it. #Day61 #RomanToInteger #LeetCode #Java #DSA #ProblemSolving #DailyCoding

#Day_63 Problem: Reverse Prefix of Word (LeetCode 2000) A clean string manipulation problem that tests attention to detail ✂️ 🧠 Insights: • Reverse only the substring till first occurrence of a character • If the character doesn’t exist, return the original string • StringBuilder makes reversal efficient ⚙️ Approach Used: • Traverse string until character is found • Reverse the prefix using StringBuilder • Append the remaining characters as-is 📌 What this improved: • StringBuilder operations • Handling early break conditions • Edge case handling ✨ Small problems sharpen precision. #Day62 #ReversePrefix #LeetCode #Java #Strings #DSA #DailyCoding

Day 38 — DSA | Linked List Cycle Detection (LeetCode 141) Today I worked on detecting a cycle in a linked list using Floyd’s Tortoise and Hare algorithm. Instead of extra memory, the idea is simple: Use two pointers Move one step vs two steps If they ever meet → cycle exists Why this problem matters: Tests pointer manipulation Reinforces thinking in O(n) time & O(1) space Very common in interviews to check fundamentals I also implemented a user-input version in Java to understand how cycles are actually formed, not just detected. Small problem, but strong concept. Consistency > complexity. #DSA #Day38 #LinkedList #Java #ProblemSolving #LeetCode #CodingJourney #FloydsAlgorithm #LearnByDoing #DataStructureAndAlgorithm

🚀 Day 50 of #100DaysOfCode Solved LeetCode Problem #1458 – Max Dot Product of Two Subsequences ✅ This problem focused on finding the maximum dot product between non-empty subsequences of two arrays. The tricky part was handling negative values and ensuring at least one pair is always chosen. Key Learnings: -> Used Dynamic Programming with memoization to avoid recomputation -> Carefully handled the base case using Integer.MIN_VALUE to enforce non-empty subsequences -> Explored the classic include vs exclude decision at each index -> Strengthened understanding of DP on two sequences Language Used: Java -> Runtime: 11 ms (Beats 59.26%) -> Memory: 48.77 MB Step by step, sharpening DP intuition and edge-case handling 🚀 #LeetCode #DynamicProgramming #Java #ProblemSolving #100DaysOfCode

#Day_59 Problem: 4Sum (LeetCode 18) Day 57 pushed the limits with 4Sum — finding all unique quadruplets equal to a target! 💥 🧠 Key Takeaways: This is an extension of 3Sum → fix two elements + two pointers. Sorting is the backbone. Duplicate skipping at both outer and inner levels is critical. Use long for sum to avoid overflow. ⚙️ Optimized Approach: Sort the array. Fix i and j. Apply two pointers for remaining part. Carefully skip duplicates. ⏱️ Complexity: O(n³), but still efficient compared to brute force. 📌 This problem taught me: How patterns scale from 2Sum → 3Sum → 4Sum. Writing clean, duplicate-free logic. Consistency + patterns = confidence! 💪 #Day57 #4Sum #LeetCode #Java #DSA #CodingPractice #LearnByDoing

LeetCode Practice - 482. License Key Formatting 🔹 Formatting Rules ✔ Remove all existing dashes (-) ✔Convert all letters to uppercase ✔Split the string into groups of size k ✔All groups must have exactly k characters ✔Except the first group, which: ✔Can be shorter than k ✔Must contain at least 1 character ✔Insert a dash (-) between each group 🔹 Why Group From Right to Left? Because: Only the first group can be smaller All other groups must be exactly k characters Grouping from the end makes this easier #LeetCode #Java #StringHandling #CodingPractice #ProblemSolving #DSA #DeveloperJourney #TechLearning