Decoding Strings with Stacks in Java

Day 64 - Intersection of Two Linked Lists Solved by using a two-pointer technique to find the intersection node of two linked lists. Instead of calculating lengths, both pointers traverse both lists to align automatically. Time Complexity: O(n + m) Space Complexity: O(1) #Day64 #LeetCode #Java #LinkedList #DSA #ProblemSolving #CodingJourney

🚀 Day 35 of my #100DaysOfCode Journey Today, I solved the LeetCode problem: Valid Anagram Problem Insight: Given two strings, check if one is an anagram of the other. Approach: • First, check if the strings have the same length; if not, return false • Convert both strings to character arrays • Sort both arrays • Compare the sorted arrays — if equal, the strings are anagrams Time Complexity: • O(n log n) — due to sorting the arrays Space Complexity: • O(n) — for the character arrays Key Learnings: • Sorting is a simple and effective way to compare character compositions • Edge cases like different lengths should be handled first • Breaking the problem into small steps makes it easy to reason about Takeaway: Sometimes, sorting can reduce a seemingly complex problem into a simple comparison. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney #ProblemSolving #Strings

✨ Day 38 of 90 – Pattern Mastery Journey 🧠 Pattern:Binary Triangle Pattern 💡 Approach: ✔ Used nested loops to control rows and columns ✔ Applied a simple condition `(i + j) % 2` to alternate values ✔ Printed ‘1’ when the sum is even, otherwise ‘0’ ✔ No extra variables needed — clean and efficient logic 🚀 This problem helped me understand how **mathematical conditions can simplify pattern logic**, making the code more optimized and readable. #PatternMasteryJourney #Java #CodingJourney #ProblemSolving

Solved a problem where we need to check if two strings can be made equal using a special operation. The rule is: you can swap characters only if the distance between their positions is even. So basically, characters at even indices can only swap among themselves, and same for odd indices. Idea: Instead of actually swapping, I just counted characters separately for even and odd positions in both strings. If both match, then it’s possible — otherwise not. Simple concept, but interesting twist! 😊 #LeetCode #DSA #Java #ProblemSolving #CodingJourney

🚀 Day 67 of #LeetCode Challenge ✅ Problem Solved: Check If Two String Arrays are Equivalent 💡 What I learned today: • Learned how to compare two string arrays without joining them • Understood how to traverse multiple strings using pointers • Improved handling of indices across arrays and strings • Realized the importance of edge cases to avoid runtime errors 🧠 Approach: • Used four pointers to track positions in both arrays and strings • Compared characters one by one • Moved to the next string when current string ends • Ensured both arrays are fully traversed at the end 📊 Key Takeaway: Efficient solutions avoid extra space — comparing character by character is better than building new strings 🔥 Consistency + small improvements every day = big progress #Day67 #LeetCode #CodingJourney #DSA #Java #ProblemSolving #Consistency

🚀 Day 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

🚀 Day 54/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1656. Design an Ordered Stream Used an array + pointer approach to store values by index and return consecutive chunks by moving a pointer forward whenever elements are available. ⏱️ Time Complexity: O(n) (amortized across all insert calls) 📦 Space Complexity: O(n) Strengthening understanding of design problems and pointer-based simulation techniques. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #CodingJourney

Mastering the Bitonic Array. 🏔️ Today's win: Finding the maximum element in an array using an optimized O(log n) Binary Search. The trick was handling the boundary conditions (checking mid == 0 or mid == n-1) to ensure the comparisons don't throw an index out of bounds. Stats: ✅ 1111 Test Cases ✅ 0.15s Runtime ✅ Total Score: 10 Seeing that green "Problem Solved Successfully" never gets old. #CodingLogic #Java #BinarySearch #BackendDevelopment #GeeksforGeeks

Day 60 - Remove Duplicates from Sorted List Worked on removing duplicate elements from a sorted linked list. Approach: • Traverse the list once • Compare current node with next node • Skip duplicate nodes by adjusting pointers Time Complexity: O(n) Space Complexity: O(1) #Day60 #LeetCode #Java #LinkedList #CodingPractice #DSA #TechJourney