Remove Element in LeetCode Challenge

🔗 DSA Practice — Generate Parentheses (LeetCode 22) Today I solved Generate Parentheses, a classic backtracking problem that beautifully demonstrates how constraints guide recursion. 🔍 Key Learnings Backtracking helps explore all valid combinations efficiently. Maintaining constraints (open vs close count) avoids invalid states. Pruning early reduces unnecessary recursive calls. Recursive tree visualization makes the logic clearer. 🧠 Why This Problem Matters One of the best problems to understand backtracking patterns. Frequently asked in coding interviews. Builds a strong foundation for combinatorial problem-solving. 📌 Final Takeaway This problem shows how powerful recursion becomes when combined with well-defined constraints. Generate only what’s valid — not everything. #leetcode #dsa #backtracking #recursion #problemSolving #codingpractice #java #interviewpreparation #algorithms #100DaysOfCode #softwareengineering

𝐃𝐚𝐲 𝟐𝟒 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on applying binary search in a 2D matrix by treating it like a flattened sorted array. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Search a 2D Matrix 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝟏 – 𝐓𝐨𝐩-𝐑𝐢𝐠𝐡𝐭 𝐒𝐜𝐚𝐧 • Start from the top-right corner • Move left if the current value is greater • Move down if the current value is smaller • Time Complexity: O(n + m) 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝟐 – 𝐑𝐨𝐰-𝐖𝐢𝐬𝐞 𝐁𝐢𝐧𝐚𝐫𝐲 𝐒𝐞𝐚𝐫𝐜𝐡 • Identify the possible row • Apply binary search inside that row • Time Complexity: O(n log m) 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝟑 – 𝐅𝐥𝐚𝐭𝐭𝐞𝐧𝐞𝐝 𝐁𝐢𝐧𝐚𝐫𝐲 𝐒𝐞𝐚𝐫𝐜𝐡 • Treat matrix as a 1D sorted array • Range: 0 to (n × m − 1) • Convert index → row = mid / m, col = mid % m • Apply standard binary search 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • A 2D matrix can be treated as 1D with index mapping • Binary search is about reducing the search space logically • Index math simplifies multidimensional problems • Choosing the right abstraction makes problems easier 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(log(n × m)) • Space: O(1) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 Sometimes the trick is not searching in 2D — it’s thinking in 1D. 24 days consistent. On to Day 25 🚀 #DSA #Arrays #BinarySearch #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

📈 DSA Practice — Check if an Array is Sorted (Recursive Approach) Today I solved Check if an array is sorted using a recursive approach, focusing on simplicity and clear base-case handling. 🔍 Key Learnings Recursion works well when a problem reduces by one element at a time. Clearly defined base cases prevent unnecessary checks. Comparing adjacent elements is enough to verify sorted order. Recursive solutions can be both clean and expressive. 🧠 Why This Problem Matters Strengthens understanding of recursion fundamentals. Commonly asked as a warm-up interview problem. Builds confidence in breaking problems into smaller subproblems. 📌 Final Takeaway This problem shows that even simple tasks benefit from structured thinking and clean recursive logic. #dsa #recursion #arrays #problemSolving #codingpractice #java #datastructures #interviewpreparation #algorithms #100DaysOfCode #softwareengineering

➕ DSA Practice — Sum of Digits (Recursive Approach) Today I solved the Sum of Digits problem using a recursive approach, focusing on breaking the problem into smaller subproblems. 🔍 Key Learnings Recursion works naturally when repeatedly reducing a number. Base case handles single-digit or zero condition. At each step, extract the last digit and recurse on the remaining number. Simple problems help strengthen core recursive thinking. 🧠 Why This Problem Matters Builds strong understanding of recursion fundamentals. Frequently asked as a basic interview or concept-check problem. Helps develop confidence in thinking step-by-step. 📌 Final Takeaway Even simple recursive problems are powerful for mastering base cases, shrinking input, and stack behavior. #dsa #recursion #mathematics #problemSolving #codingpractice #java #datastructures #interviewpreparation #algorithms #100DaysOfCode #softwareengineering

🚀 DSA Insight — Reversing an Array the Optimal Way A very common problem, but also a great test of fundamentals. Many beginners create a new array or use extra memory to reverse elements… But the most efficient solution doesn’t require any extra space 👇 🏆 Optimal Approach — Two Pointer Technique Use two pointers: left → start of array right → end of array Swap elements and move inward. swap(arr[left], arr[right]) left++ right-- We stop when pointers meet in the middle. 📊 Complexity MetricValueTime ComplexityO(n)Space ComplexityO(1)Swapsn/2💡 Why It’s Optimal To reverse an array, every element must move at least once → O(n) is unavoidable. This approach also avoids extra memory, making it optimal in both time and space. 🧠 Simple logic + strong fundamentals = efficient coding #DSA #Algorithms #CodingInterview #Java #ProblemSolving #SoftwareEngineering

Day 18/100 of DSA , (Arrays) 🚀Toady I Solved: classic 3Sum problem — and it turned out to be more about thinking than just coding. 💡 At first, brute force (three nested loops) seemed simple. But O(n³) complexity clearly isn’t scalable. 🚫 Here’s what I applied: 🔹 Sorted the array 🔹 Fixed one element at a time 🔹 Used the Two Pointer technique 🔹 Carefully handled duplicates 🔹 Optimized time complexity to O(n²) What I learned today: Problem solving isn’t about memorizing answers. It’s about training your brain to recognize patterns and make logical decisions under constraints. 🧠 Small improvements in logic create big improvements in performance. 📈 Every DSA problem I practice is a step toward stronger fundamentals and better interview confidence. 🚀 #DSA #Java #ProblemSolving #InterviewPreparation #CodingJourney #FutureDeveloper

Daily DSA Update – Day 26 Solved: Add Binary (LeetCode) Today’s problem was about adding two binary strings and returning their sum as a binary string. Problem: Given two binary strings a and b, return their sum as a binary string. Approach: I simulated the same process we use while doing binary addition manually. Starting from the end of both strings, I added corresponding digits along with a carry value. After calculating the sum, I appended the result bit and updated the carry. Finally, I reversed the result to get the correct binary order. Key Learning: Problems like this highlight the importance of understanding how operations work internally rather than relying on built-in conversions. What this strengthened: • String traversal from right to left • Handling carry in binary operations • Using StringBuilder for efficient string manipulation • Implementing mathematical logic in code Each daily problem adds a small but meaningful improvement in problem-solving ability. On to the next challenge. #DSA #DataStructures #Algorithms #Java #LeetCode #ProblemSolving #CodingJourney #TechLearning

Daily DSA Update – Day 25 Solved: Length of Last Word (LeetCode) Today’s problem looked simple at first glance, but it reinforced an important lesson — clean logic beats shortcut methods. Problem: Given a string containing words and spaces, return the length of the last word. Approach: Instead of using split() (which creates extra space), I traversed the string from the end: • Skipped trailing spaces • Counted characters until the next space appeared Why this matters: Avoiding unnecessary string operations improves space efficiency and shows better control over logic implementation. Key Learning: Sometimes the most optimal solution is not about complex algorithms but about handling edge cases carefully — especially trailing spaces and empty inputs. What this strengthened: • String traversal techniques • Edge case handling • Writing space-optimized code • Thinking from an interviewer’s perspective Consistency is slowly building clarity in problem solving. On to the next challenge. #DSA #DataStructures #Algorithms #Java #LeetCode #ProblemSolving #CodingJourney #DailyLearning

Day 14/100 | #100DaysOfDSA Today’s problem: Search a 2D Matrix At first glance, it looks like a 2D search problem… But the trick? Treat the matrix like a sorted 1D array. Since: • Each row is sorted • First element of every row > last element of previous row That means the entire matrix is globally sorted. Approach: • Apply Binary Search • Convert mid index → row = mid / n, col = mid % n • Compare and adjust pointers Time complexity: O(log(m × n)) Big takeaway: Sometimes the problem isn’t 2D at all — it’s just perspective. Stacking binary search patterns now. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearch #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity #SoftwareEngineer #ComputerScience #Consistency #Programmers #TechGrowth

🚀 Problem Solved: Pow(x, n) – Fast Exponentiation Today I worked on LeetCode 50 (Medium) — implementing pow(x, n) efficiently. Instead of using the naive O(n) multiplication approach, I applied Binary Exponentiation (Fast Power Algorithm) to reduce the time complexity to O(log n). 🔎 Key Learnings: Handling negative powers correctly Preventing integer overflow (using long for edge cases) Understanding how dividing the exponent by 2 optimizes performance Applying mathematical logic in coding interviews Example: 2¹⁰ → 1024 2⁻² → 0.25 This problem was a great reminder that optimization isn’t optional — it’s essential. Slowly building strong fundamentals, one problem at a time. 💻✨ #LeetCode #ProblemSolving #Java #DataStructures #Algorithms #CodingJourney #InterviewPrep #BinaryExponentiation