Linked List Manipulation: Move Last Node to Front

🚀 Day 40/100 LeetCode Challenge Completed! Problem Solved: Minimum Absolute Difference Today’s challenge involved finding all pairs of elements in an array with the minimum absolute difference. This problem tests your understanding of array manipulation, sorting, and efficient traversal. 🔍 Approach & Solution Highlights: Sorting First: By sorting the array initially, we ensure that adjacent elements are closest in value, simplifying the difference calculations. Single Pass Traversal: After sorting, we traverse the array once to compute differences between consecutive elements. Dynamic Result Building: Instead of storing all pairs and filtering later, we maintain the current minimum difference and update the result list on the fly. Time Complexity: O(n log n) due to sorting, with O(n) traversal. Space Complexity: O(1) extra space (excluding output storage). 💡 Key Takeaways: Sorting as a Preprocessing Step is a powerful technique to simplify problems involving differences or distances between elements. Efficient State Management—clearing and rebuilding the result list only when a new minimum is found—avoles unnecessary storage and operations. Edge Cases Handled: Works seamlessly with both positive and negative integers, and handles arrays of varying lengths as per constraints. 📈 Why This Matters: Problems like these are common in technical interviews, especially for roles involving data analysis, algorithm design, and optimization. They demonstrate your ability to think step-by-step, optimize runtime, and write clean, efficient code. 🧩 Example Walkthrough: Input: [4,2,1,3] Sorted: [1,2,3,4] Differences: [1,1,1] Minimum Difference: 1 Pairs: [[1,2],[2,3],[3,4]] This solution ensures we only return valid pairs in ascending order, meeting all given constraints. #LeetCode #CodingChallenge #100DaysOfCode #Algorithm #DataStructures #ProblemSolving #Java #SoftwareEngineering #Tech #Developer #CodingLife #InterviewPrep #TechCommunity #CodeNewbie #Programming #DailyCoding #LearnToCode #SoftwareDevelopment #ComputerScience

Optimization Matters: How I solved LeetCode 1390 (Four Divisors) 🚀 When solving algorithmic problems, the difference between a "working" solution and an "accepted" solution usually comes down to one thing: Time Complexity. The Challenge: The problem asks for the sum of divisors of numbers that have exactly four divisors. With values up to 10⁵and an array size of 10⁴, a brute-force approach (checking every number from 1 to n for every element) would result in O(N×M) complexity. In the worst case, that's 10⁹ operations—way too slow for most environments. My Evolution of Logic: Initially, I considered a simple linear scan for divisors. However, I realized I needed a more mathematical approach to stay within the time limits. The Optimized Approach: I shifted to a Square Root of x optimization. Here’s why it works: Divisor Pairs: Divisors always appear in pairs. For a number x, if i is a divisor, then x/i is also a divisor. One will always be < √xand the other will be >=√x Efficiency: By only iterating up to √x, I reduced the operations per number from 100,000 to just 316. The Logic: Loop from 1 to √x If (x% i == 0), identify if it’s a perfect square (i == x/i) or a distinct pair. Early Exit: If the divisor count exceeds 4, I immediately break the loop. This "pruning" saves significant CPU cycles. Only if the final count is exactly 4, do I add the sum to the total. Key Takeaway: Writing code that works is the first step, but writing code that is performant is the goal. This approach brought the complexity down to O(N√M}), making the solution run almost instantly. #SoftwareEngineering #Coding #Optimization #Java #LeetCode #DSA #ContinuousLearning #DailyCoding #LeetCodeChallenge

🚀LeetCode POTD | 1st Jan 2026 Plus One (LC 66) | Consistency > Intensity ✅ Kicking off 2026 with a classic that quietly tests your fundamentals. Plus One looks simple—but it’s a sharp reminder that edge cases separate clean code from broken logic. 🔹 Core Idea You’re given a number as an array of digits. Add +1 and handle carry like a pro. 🔹 What This Problem Actually Teaches ✔ How to process arrays from right to left ✔ Handling carry without converting to integers ✔ Writing space-efficient, in-place logic ✔ Thinking about worst cases (all 9s 👀) 🔹 Why It Matters This pattern shows up everywhere: big numbers, string math, system constraints, and interview rounds where “easy” questions aren’t really easy. 💡 Takeaway Mastering basics isn’t optional—it’s compound interest for your problem-solving skills. Day 1. Momentum started. On to the next one. 🔥 #LeetCode #LeetCodePOTD #DSA #DataStructures #Algorithms #Java #JavaDeveloper #ProblemSolving #CodingInterview #InterviewPrep #SoftwareEngineering #CompetitiveProgramming #CleanCode #TimeComplexity #SpaceComplexity #Arrays #LogicBuilding #TechCareers #Developers #DailyCoding #Consistency #LearningInPublic #CodingJourney #2026Goals #NewYearNewSkills

🔥 LeetCode Daily | Hard Problem Cracked (1458: Max Dot Product of Two Subsequences) 🔥 Another day, another reminder that real problem-solving begins where shortcuts fail. Today’s challenge wasn’t about syntax or speed — it was about decision-making under constraints. With negative numbers in play, greedy approaches collapse fast. The only way forward? Clean, well-structured Dynamic Programming. 🧠 What made this problem interesting: • Choosing take vs skip at every index • Handling negative values without breaking the logic • Ensuring subsequences remain non-empty (the real twist) ⚙️ Approach: Dynamic Programming to track the best possible dot product at each state while preserving order and maximizing value. 📊 Complexity: • Time: O(n × m) • Space: O(n × m) 💡 Lesson learned: Hard problems don’t test how fast you code. They test how clearly you think when the obvious approach fails. Consistency + fundamentals = inevitable progress. On to the next challenge. 🚀 #LeetCode #LeetCodeDaily #HardProblems #DynamicProgramming #DP #DataStructures #Algorithms #DSA #ProblemSolving #CodingInterview #InterviewPreparation #CompetitiveProgramming #Java #JavaDeveloper #SoftwareEngineering #ComputerScience #TechCareers #DeveloperJourney #100DaysOfCode #DailyCoding #ConsistencyWins

🚀 𝑫𝒂𝒚 7 𝒐𝒇 𝑴𝒚 𝑳𝒆𝒆𝒕𝑪𝒐𝒅𝒆 𝑱𝒐𝒖𝒓𝒏𝒆𝒚 | 𝑫𝑺𝑨 – 𝑺𝒕𝒂𝒄𝒌 (𝑱𝒂𝒗𝒂) Continuing my daily DSA practice with LeetCode 📘 Today was all about using stacks to solve real-world array problems efficiently and understanding the concept of Next Smaller Element. 🔹 𝑷𝒓𝒐𝒃𝒍𝒆𝒎 𝑺𝒐𝒍𝒗𝒆𝒅 𝑻𝒐𝒅𝒂𝒚:   𝑭𝒊𝒏𝒂𝒍 𝑷𝒓𝒊𝒄𝒆𝒔 𝑾𝒊𝒕𝒉 𝒂 𝑺𝒑𝒆𝒄𝒊𝒂𝒍 𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕 𝒊𝒏 𝒂 𝑺𝒉𝒐𝒑 (https://lnkd.in/gNZSzHNQ) 🔹 Key Concept Used: Monotonic Stack (Next Smaller or Equal Element to the Right) 🔹 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 𝐀𝐧𝐚𝐥𝐲𝐬𝐢𝐬: Time Complexity: O(n) Space Complexity: O(n) 🔹 What I’m Learning: • Stack helps avoid unnecessary nested loops • Index-based stack usage simplifies discount calculation • Recognizing patterns like “next smaller element” is crucial • Clean logic + practice = faster problem solving Solving one problem a day and strengthening fundamentals step by step. Consistency over intensity.🚀 #LeetCode #DSA #Stacks #Java #ProblemSolving #DailyCoding #Consistency #LearningJourney #BackendDeveloper #2026Goals

🎯 #LeetcodePotd | Problem Solved with 99%+ Runtime Performance Sometimes the smartest solution isn’t fancy — it’s structured thinking: Just solved LeetCode 1984 – Minimum Difference Between Highest and Lowest of K Scores ✅ And honestly? This is a perfect example of how sorting + sliding window = instant clarity. 🧠 Problem in Simple Words You’re given student scores. Pick k students such that the difference between the highest and lowest score is as small as possible. 👉 Goal: Minimize (max − min) among any group of size k. 💡 Key Insight (Beginner Friendly) If the array is sorted, then: Any group of k students will be contiguous The difference becomes: nums[i + k − 1] − nums[i] So instead of brute force chaos, we: Sort the array Slide a window of size k Track the minimum difference Clean. Efficient. Scalable. ⏱️ Complexity Analysis Time Complexity: O(n log n) (sorting dominates) Space Complexity: O(1) (ignoring sorting internals) 📊 Submission Stats ✔ Accepted: 118 / 118 test cases ⚡ Runtime: 7 ms 🔥 Beats 99.17% of submissions 💾 Memory: 47.15 MB 🧠 Takeaway Sort first. Think visually. Then optimize. On to the next one. Consistency > motivation. 💪 #LeetCode #DSA #Java #ProblemSolving #Algorithms #SlidingWindow #Sorting #CodingJourney #BeginnersInTech #100DaysOfCode #SoftwareEngineering #InterviewPrep

Day 26/100 of the LeetCode Challenge — Problem Solved! ✅ Today's problem presented an interesting twist on movement in a 2D grid: given a sequence of points, find the minimum time to visit them in order while moving horizontally, vertically, or diagonally in 1 second per unit step. At first glance, this could invite thoughts of BFS, Dijkstra, or path optimization techniques. But a closer look reveals a clean mathematical insight — the optimal path between two points leverages diagonal movement as much as possible, which essentially covers both horizontal and vertical distance simultaneously. The key realization? The minimal time between (x1, y1) and (x2, y2) is simply max(|dx|, |dy|). You move diagonally for min(|dx|, |dy|) seconds, covering both axes at once, and then move straight along the longer axis for the remaining difference. So the overall solution becomes beautifully simple: Iterate through consecutive points Compute absolute differences in x and y Take the maximum of the two Sum them all up What appeared to be a search or simulation problem turned into a one-liner of arithmetic per step. Why this matters: In both coding interviews and real-world system design, recognizing when to apply a direct formula instead of a heavy algorithm is crucial. It reflects clarity of thought, mathematical intuition, and the ability to simplify without over-engineering. On to the next challenge! 🚀 #HemaliCodes #LeetCode #100DaysOfCode #CodingChallenge #Algorithm #ProblemSolving #Tech #SoftwareEngineering #Java #WomenWhoCode #Developer #Programming #TechJourney #CodeNewbie #DailyCoding #LearnInPublic #TechnicalSkills #InterviewPrep #MathInCode

🚀 Day 32 of LeetCode Problem Solving Journey #100DaysOfLeetCode Challenge: Today, I solved the problem: "Final Value of Variable After Performing Operations". It is an Easy-level problem that tests basic string handling and logic building. Problem Explanation: There is a programming language with only four operations and one variable X (initially 0): ++X and X++ increment the value of the variable X by 1. --X and X-- decrement the value of the variable X by 1. Given an array of strings containing a list of operations, we need to return the final value of X after performing all of them. Example: Input: operations = ["--X","X++","X++"] Output: 1 (Explanation: 0 - 1 + 1 + 1 = 1) My Approach & Logic: I solved this using a simple iteration method. Here is the breakdown: 🔹 Logic Breakdown: Initialization: I started with an integer variable ans initialized to 0 to keep track of the value of X. Iterating through Operations: I used a range-based for loop to traverse the input vector operations. Checking Conditions: Inside the loop, I checked if the current operation string was an increment operation ("X++" or "++X"). If true, I incremented ans by 1. Otherwise (since the valid inputs are limited), I treated it as a decrement operation and decreased ans by 1. Result: Finally, I returned the stored value in ans. This approach processes the operations efficiently in a single pass! Big thanks to NEKAL SINGH SALARIA sir at REGex Software Services for the continuous mentorship and guidance. On to Day 33! 💻 #100DaysLeetCode #Day32 #100DaysOfCode #CPP #Strings #Coding #LogicBuilding #ProblemSolving #DSA #DataStructures #CodingChallenge #LeetCode #Mentorship #ConsistencyIsKey #REGexSoftwareServices REGex Software Services

𝗗𝗮𝘆 𝟮𝟳/𝟭𝟬𝟬 | 𝗗𝗲𝗰𝗼𝗱𝗲 𝗦𝘁𝗿𝗶𝗻𝗴 Day 27 ✅ — When recursion makes everything elegant. 𝗧𝗼𝗱𝗮𝘆'𝘀 𝗣𝗿𝗼𝗯𝗹𝗲𝗺: ✅ 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 #𝟯𝟵𝟰: Decode String (Medium) 𝗪𝗵𝗮𝘁 𝗖𝗹𝗶𝗰𝗸𝗲𝗱: Decode strings like "3[a2[c]]" → "accaccacc". Numbers indicate repetition, brackets show nested patterns. My first thought? Stack. But then I realized: nested brackets = 𝗿𝗲𝗰𝘂𝗿𝘀𝗶𝗼𝗻 territory. Each time I hit '[', recursively decode what's inside. When I hit ']', return to the outer level. The recursion naturally handles the nesting. 𝗠𝘆 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵: 👉 Build number from consecutive digits 👉 When '[' appears, recursively decode inner string 👉 Repeat the decoded string by the number 👉 When ']' appears, return result to caller 👉 Regular characters get appended directly Time: O(maxK × n) where maxK is max repeat count, Space: O(n) for recursion stack 𝗠𝘆 𝗥𝗲𝗮𝗹𝗶𝘇𝗮𝘁𝗶𝗼𝗻: Recursion isn't just for trees and graphs—it's for any problem with 𝗻𝗲𝘀𝘁𝗲𝗱 𝘀𝘁𝗿𝘂𝗰𝘁𝘂𝗿𝗲𝘀. The key: recognize when a subproblem looks exactly like the main problem. That's your recursion signal. 27 days in, and I'm getting more comfortable with recursive thinking. 𝗖𝗼𝗱𝗲: https://lnkd.in/gFfUH6bk 𝟮𝟳 𝗱𝗮𝘆𝘀 𝘀𝘁𝗿𝗮𝗶𝗴𝗵𝘁. 𝗔𝗹𝗺𝗼𝘀𝘁 𝗮 𝗺𝗼𝗻𝘁𝗵. Every problem teaches a pattern. Every pattern becomes a tool. 𝗗𝗮𝘆 𝟮𝟳/𝟭𝟬𝟬 ✅ | 𝟳𝟯 𝗺𝗼𝗿𝗲 𝘁𝗼 𝗴𝗼! #100DaysOfCode #LeetCode #Recursion #StringManipulation #DataStructures #CodingInterview #TechnicalInterview #SoftwareEngineer #FAANG #ProblemSolving #Algorithms #Java #Programming #ComputerScience #DeveloperJourney #LearnInPublic #SoftwareDevelopment #CodingSkills #TechJobs #CareerGrowth