LeetCode Challenge: Check if Digits Are Equal After Operations

#DSAChallenge | #LeetCodeProgress 📌 Problem: 54.Spiral Matrix 📍 Platform: LeetCode Over the past few days, I’ve been exploring some classic yet challenging LeetCode problems — the kind that really test your understanding of logic, edge cases, and clean code design. 🧠 Problem Insight You're given a matrix, and the goal is to return all elements in spiral order — starting from the top-left corner and moving right → down → left → up, repeating this pattern until all elements are covered. ⚙️ Logic Breakdown (Step-by-Step): 1️⃣ Set boundaries: top, bottom, left, right 2️⃣ Traverse in order: → Move left to right (top row) → top++ → Move top to bottom (right column) → right-- → Move right to left (bottom row) → bottom-- → Move bottom to top (left column) → left++ 3️⃣ Continue while top <= bottom and left <= right 🧩 Key Learning Points: ✅ Strengthened my understanding of matrix boundaries and directions. ✅ Improved my ability to handle edge cases cleanly. ✅ Learned how to manage loop conditions efficiently. ⏱️ Time & Space Complexity Time Complexity: O(m × n) → Every element is visited once Space Complexity: O(1) → If output list is excluded 💡 Takeaway: In DSA, consistency beats intensity. Even small daily challenges — when done consistently — compound into stronger problem-solving intuition. 🔥 Next Up: Moving on to Rotate Image & Set Matrix Zeroes to deepen my matrix manipulation skills! #DSA #LeetCode #ProblemSolving #DataStructures #Algorithms #100DaysOfDSA #Matrix #SpiralMatrix #CodeNewbie #GrowthMindset #DailyLearning #NeverGiveUp #KeepCoding #LearningNeverStops

🚀 LeetCode Daily Challenge Complete! Just solved "Final Value of Variable After Performing Operations" - a simple yet elegant problem that teaches the value of finding shortcuts in pattern matching! 💡 Solution Approach: ✅ Iterate through all operation strings ✅ Check middle character (index 1) ✅ If '-': decrement, if '+': increment ✅ Return final value The key insight: Instead of comparing full strings ("--X", "X--", "++X", "X++"), we can observe that ALL operations have the operator in the middle position! By checking just op[1], we instantly know whether to increment or decrement. Pattern recognition: "--X" and "X--" both have '-' at index 1 → decrement "++X" and "X++" both have '+' at index 1 → increment This single-character check is more efficient than string comparisons! Time Complexity: O(n) | Space Complexity: O(1) #LeetCode #StringParsing #CPlusPlus #Algorithms #SoftwareEngineering #ProblemSolving #CleanCode #PatternRecognition

🚀 LeetCode Challenge: Sqrt(x) Today’s problem was all about finding the square root of a number without using any built-in functions , sounds simple, but implementing it from scratch really makes you think deeper! After doing some research online, I discovered that this problem can be efficiently solved using Binary Search . Binary Search isn’t just for finding elements in a sorted array , it’s a technique that helps you narrow down a search space efficiently. 👉 Here’s the logic in simple words: The square root of x always lies between 0 and x. Instead of checking every number, we can repeatedly divide this range in half. If mid² is greater than x, the answer lies on the left side. If mid² is smaller, we move right , because we need a larger number. And if it’s equal , that’s our perfect square root. By updating the range step by step, we finally land at the integer part of the square root , with much better efficiency. 💡This problem wasn’t about math at all , it was about thinking efficiently. Sometimes in coding (and in life), we waste time checking every option linearly when we could just “binary search” our way , observe, eliminate, and focus on what matters. P.S. It’s my fourth day of LeetCode problem-solving. Here’s my LeetCode profile 👇 https://lnkd.in/daSxpGmi 🔗 Connect if you are also solving LeetCode problems — I’ll be sharing simple explanations of the problem solutions here. 🔁If you think this could help another learner, share it forward #datastructures #algorithms #LeetCode #squarerootofx #sqrt(x) #leetcodeproblem69

#28 🚀 Problem Solved: Broken Calculator (LeetCode) 💻 Today I tackled the "Broken Calculator" problem — an interesting challenge that sharpens your greedy algorithm skills! 🧩 Problem Summary: Given two integers startValue and target, the goal is to determine the minimum number of operations needed to make startValue equal to target. The allowed operations are: 1️⃣ Multiply by 2 2️⃣ Subtract 1 🔍 Key Insight: Instead of moving from startValue → target, it’s more efficient to work backwards — from target → startValue. If target is even, divide it by 2; if odd, add 1. Continue until target <= startValue. Finally, add the difference to account for the remaining steps. ✅ Complexity: Time: O(log target) Space: O(1) Solving this reinforced how thinking in reverse can simplify what seems like a complex forward process — a useful mindset for both algorithms and life! 🌱 #LeetCode #Coding #ProblemSolving #Algorithms #C++ #GreedyAlgorithm #LearningEveryday

Day 5/365 — Tackling Linked Lists & Recursion 🚀 Another milestone on my LeetCode journey! Today, I solved five classic problems that deepened my understanding of linked lists and recursion. Problems Solved Today: 21. Merge Two Sorted Lists 22. Generate Parentheses 23. Merge k Sorted Lists 24. Swap Nodes in Pairs 25. Reverse Nodes in k-Group Today's Stats: 🕒 Time spent: about 3 hours 🧠 Concepts used: linked list manipulation, dummy nodes, recursion, divide and conquer, iterative/pointer techniques ✨ Biggest insight: Attacking complex list problems with both recursion and iterative logic—and using dummy nodes—makes tricky edge cases much easier to handle. 3 Tips Learned: 1. Using a dummy node can simplify pointer management for list merges. 2. Break down recursive problems like parentheses/gen. into base case + valid subparts. 3. For k-group reversal, visualize node connections before coding to avoid pointer mishaps. Why this challenge? Every new pattern mastered compounds into long-term algorithmic confidence. Let's keep grinding! Are you on a similar journey? Drop a 👍 if you're committed too, and share your favorite coding tip! #LeetCode #365DaysOfCode #Algorithms #DataStructures #ProgrammingJourney #ProblemSolving #GrowthMindset #LearningInPublic #DeveloperLife

Solving the Rotated Challenge: Search in Rotated Sorted Array (LeetCode 33)! 🚀 You know Binary Search is O(\log N), but what happens when the array is sorted and rotated? The standard two-pointer approach breaks down! The key insight is to realize that even though the whole array is rotated, at least one half of the array must be perfectly sorted. We exploit this property in every iteration. The Modified O(\log N) Approach: Find the midpoint, \text{mid}. Identify the Sorted Half: Check if \text{nums}[\text{left}] \le \text{nums}[\text{mid}]. If true, the left half is sorted. Otherwise, the right half is sorted. Target Check: If the \text{target} falls within the boundaries of the sorted half, we discard the other half. If the \text{target} is not in the sorted half, it must be in the unsorted half, so we discard the sorted half. This technique ensures we always cut the search space in half, maintaining the optimal O(\log N) time complexity. Always a satisfying problem to solve! #Algorithms #BinarySearch #CodingChallenge #LeetCode #TechSkills

🔹 Day 67 of #100DaysOfLeetCodeChallenge 🔹 Problem: Combination Sum Focus: Backtracking + Recursion 💡 The Challenge: Find all unique combinations of numbers that sum to a target. The twist? You can reuse the same number unlimited times! This makes it more interesting than standard subset problems. 🧠 My Approach: Used backtracking with two key decisions at each step: Pick: Include current element and explore further (stay at same index for reuse) Skip: Move to next element without including current one Base case: When we've explored all elements, check if target reached 0 Optimization: Only pick if element ≤ remaining target 📊 Complexity Analysis: ⏳ Time: O(2^t) where t = target/min(candidates) — worst case recursion tree 💾 Space: O(t) — maximum recursion depth 📌 Example: Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: - 2+2+3 = 7 (reused 2 twice!) - 7 = 7 (single element) 🎯 Key Takeaway: The beauty of this problem lies in handling unlimited reuse. By staying at the same index when picking an element, we elegantly allow repetition without creating duplicates. This pick/skip pattern is fundamental to combinatorial backtracking! Real-world analogy: Like making change for a dollar — you can use the same coin denomination multiple times! 💰 Day 67/100 complete. Mastering backtracking, one problem at a time! 💪 #LeetCode #Backtracking #DynamicProgramming #DSA #Algorithms #CodingInterview #100DaysOfCode #SoftwareEngineering #ProblemSolving #TechCareer

Back on the Grind: Tackling "Find Median from a Data Stream" After a short hiatus, I'm thrilled to be diving back into algorithmic problem-solving! There's no better way to restart than with a classic: LeetCode 295. Find Median from Data Stream. This problem is a fantastic showcase of how choosing the right data structures is everything. The challenge is to design a data structure that supports: Adding integers. Finding the median efficiently at any time. A naive approach might involve re-sorting, but that's inefficient. The elegant solution? The Two-Heap Pattern. The Insight: By using a max-heap for the lower half of the numbers and a min-heap for the upper half, we can always access the middle values (the medians) in constant time, O(1). Insertions are handled efficiently in O(log n) by balancing the heaps. Why I love this problem: It’s a perfect reminder that complex problems often have beautiful and efficient solutions. It reinforces core concepts that are crucial for system design and coding interviews. Re-committing to a consistent practice schedule. On to the next one! What's a recent problem you've solved that you found particularly insightful? #LeetCode #Algorithm #DataStructures #ProblemSolving #SoftwareDevelopment #Coding #Tech #Programming #CareerGrowth #Consistency

🚀 Ever wondered how we actually measure how good an algorithm is? There are two ways — and understanding the difference can instantly make you a smarter developer 👇 💡 1. A Priori Analysis (Before Execution) This is all about theory. You don’t run the code — you just analyze its logic. For example, you can tell that Bubble Sort takes O(n²) time while Merge Sort takes O(n log n), just by studying their steps. It’s like predicting how long a trip will take before actually driving. ⚙️ 2. A Posteriori Testing (After Execution) This is all about practice. You actually run the code, measure how long it takes, and record memory usage. For example, you might find Bubble Sort takes 15 seconds for 10,000 numbers, while Merge Sort finishes in 0.01 seconds. That’s real-world evidence. 📊 In short: A Priori = Predictive (mathematical reasoning) A Posteriori = Experimental (actual performance) Both are essential: 👉 A Priori tells you how it should perform. 👉 A Posteriori tells you how it actually performs. Understanding both sides of the equation helps us build faster, smarter, and more efficient systems 🔥 #Coding #Algorithms #ComputerScience #DSA #LearningEveryday #DeveloperLife

🚀Day 77 of #120_Days_leetCode Challenge ! LeetCode Problem: Find and Replace Pattern Today’s problem was all about pattern matching with bijective mappings — where we find words that match a given pattern by establishing a one-to-one relationship between letters. 🧩 Problem Summary: Given a list of words and a string pattern, return all words that match the pattern. A word matches if you can permute letters such that each letter in the pattern maps uniquely to a letter in the word. 📘 Example: Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb" Output: ["mee","aqq"] 💡 Intuition: To verify a match, we need to ensure: Each character in the pattern maps to only one unique letter in the word. No two pattern characters map to the same word character. This makes it a bijection problem — a perfect use case for hash maps. 🧠 Key Takeaways: Used two hash maps to maintain bijective character mapping. Achieved O(n * k) time complexity, where n = number of words and k = word length. Great practice for mastering string and hashmap-based logic problems. ✨ Every such problem reinforces how powerful simple mapping logic can be when applied thoughtfully. #LeetCode #ProblemSolving #CodingChallenge #DataStructures #Algorithms #ProgrammingJourney