How LeetCode 221 taught me discipline over syntax

📌 Day 155 of Coding - Next Greater Numerically Balanced Number (LeetCode - Medium) 🎯 Goal: Find the smallest integer greater than n such that the count of each digit d in the number is exactly d (for digits 1-7). 💡Approach & Debugging: Used backtracking to generate all “beautiful” numbers following the digit count rule. A helper function generate() recursively built numbers by trying digits 1-7, ensuring that no digit appeared more times than its own value. Checked each generated number using isBeautiful(). After generation, sorted the list and returned the first number greater than n. ✔️ Time Complexity: Exponential (backtracking-based generation, but limited by small constraints). ✔️ Space Complexity: O(1) auxiliary + O(M) list for generated numbers. 🧠 Key Takeaways: Precomputation and pruning (like limiting to 1224444) help keep recursion efficient. Problems mixing digit frequency logic and recursion sharpen number theory intuition. #Day155 #LeetCode #Backtracking #Recursion #Java #ProblemSolving #DSA #CodingChallenge #Algorithms #100DaysOfCode #InterviewPrep #SoftwareEngineering #DataStructures #CodeNewbie #ProgrammersLife

🚀 Day 40 of My #LeetCode Challenge 🚀 📘 Problem: Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold (LeetCode #1343) 💡 Approach: Implemented the Sliding Window technique to efficiently calculate the sum of each subarray of size k. First, compute the sum of the initial k elements. Then, slide the window forward — subtract the element leaving and add the new one entering. Check if the average of the current window meets or exceeds the given threshold. This method optimizes the time complexity to O(n) by avoiding repetitive summation. ⏱️ Concepts Used: Sliding Window | Arrays | Optimization | Time Complexity 🔥 Consistency and logic go hand in hand — every day adds a new layer to my problem-solving skills! #Day40 #LeetCode #100DaysOfCode #CodingChallenge #Java #DSA #ProblemSolving #CodingJourney #TechLearning #CodeEveryday #SoftwareEngineering #ProgrammerLife #CodingCommunity #Developer #JavaProgramming #LearnToCode #TechGoals #CodeNewbie #Algorithm #DataStructures #CodingPractice

🚀 Day 68 of #100DaysOfChallenge Today's problem: 3461. Check If Digits Are Equal in String After Operations I (LeetCode - Easy) This problem was a fun blend of pattern observation and modular arithmetic — a great reminder that even simple-looking questions can strengthen your logical flow and coding discipline on LeetCode. 🧩 Problem Overview: We’re given a string of digits, and in each step, we repeatedly calculate the sum of every pair of consecutive digits modulo 10, until only two digits remain. Finally, we check whether those two digits are equal. 💡 My Approach: Used a straightforward simulation approach — repeatedly computing new strings of digits using modular arithmetic until only two digits were left. It’s simple, direct, and performs efficiently for short strings. ⚙️ Language: Java ⚡ Runtime: 8 ms (Beats 72.03%) 💾 Memory: 44.67 MB (Beats 69.77%) ✅ Result: Accepted — 706 / 706 test cases passed Every daily problem adds another layer of precision and confidence. Consistency isn’t just about coding every day — it’s about learning something new with every attempt. 🌱 On to the next challenge 💪 #100DaysOfCode #LeetCode #ProblemSolving #Java #CodingChallenge #Programming #DeveloperJourney #LearningEveryday #TechMindset #Consistency

📌 Day 161 of Coding - Minimum Number of Increments on Subarrays to Form a Target Array (LeetCode - Hard) 🎯 Goal: Find the minimum number of operations required to form a target array starting from an array of zeros, where each operation increments all elements of a chosen subarray by 1. 💡Approach & Debugging: Used a difference-based greedy approach to count only the necessary increments. Steps: Initialize sum with the first element, representing the increments needed for the first number. Traverse the array: Whenever target[i] > target[i - 1], it means additional increments are needed. Add the difference (target[i] - target[i - 1]) to sum. The total sum gives the minimum number of operations required. ✔️ Time Complexity: O(N) - single pass through the array. ✔️ Space Complexity: O(1) - constant extra space. 🧠Key Takeaways: A classic prefix difference trick, focus on the increase, not the absolute value. Greedy approaches often emerge naturally in array transformation problems. #Day161 #LeetCode #GreedyAlgorithm #ArrayProblems #Java #ProblemSolving #DSA #Algorithms #CodingChallenge #InterviewPrep #100DaysOfCode #SoftwareEngineering #CodeNewbie #ProgrammersLife

🚀 Day 8 of #45DaysOfLeetCode Challenge 😎 📌Today's problem: Palindrome Number (LeetCode #9) 💡 🔹 Concept: Check whether a given integer reads the same backward and forward — without converting it to a string! 🔹 Logic Used: Instead of reversing the entire number, I reversed only half of it to improve efficiency. This avoids unnecessary computation and eliminates integer overflow risks. 🔹 Key Learnings: ✅ Optimized logic using mathematical manipulation ✅ Improved understanding of number reversal techniques ✅ Importance of edge case handling (negative numbers, trailing zeros) ⚙️ Result: 💻 Runtime: 4 ms (Beats 100%) 💾 Memory: 44.84 MB 😎Small optimizations make a big difference in performance! 💪 📌Problem: https://lnkd.in/ef6AC2j6 🔥Each day is a step closer to writing cleaner and more optimized code. ✨ Let’s keep pushing forward and refining our problem-solving skills! 💻🔥 #LeetCode #Day8 #100DaysOfCode #ProblemSolving #Java #CodingChallenge #PalindromeNumber #DataStructures #Algorithms #LearningEveryday

💥 Stop solving 500 random Leetcode questions! If you can master these 7 patterns, you can crack any DSA problem. Most developers don’t need more problems they need clarity on the patterns behind them. 🚀 Here’s what I wish I knew earlier: If you understand 1. Sliding Window 2. Two Pointers 3. Binary Search 4. Recursion 5. Dynamic Programming 6. Greedy 7. Backtracking ...then every new problem will start to look familiar. 💡 Save this post 🔖 and start with one pattern a week. Your consistency will matter more than your question count. 👇 Comment which pattern took you the longest to understand! #DSA #Java #BackendDevelopment #Leetcode #ProgrammersLife

✅ Day 59 of LeetCode Medium/Hard Edition Today’s challenge was “Maximum Alternating Subsequence Sum” — a beautifully balanced dynamic programming problem that turns simple addition into an elegant alternation of signs ⚖️➕➖ 📦 Problem: Given an integer array nums, you need to choose a subsequence (not necessarily contiguous) that maximizes its alternating sum, defined as the sum of elements at even indices minus the sum of elements at odd indices. For example: nums = [4,2,5,3] → optimal subsequence [4,2,5,3] gives (4 + 5) - (2 + 3) = 4. 🔗 Problem Link: https://lnkd.in/g5M-iugB ✅ My Submission: https://lnkd.in/gUmnEyUU 💡 Thought Process: At each index, we have two choices — either include the current element (flipping the sign based on its position in the subsequence) or skip it. This naturally forms a two-state dynamic programming model: turn = 0 → we add the number (even position) turn = 1 → we subtract the number (odd position) The recurrence relation becomes: dp[i][turn] = max( (turn == 0 ? +nums[i] : -nums[i]) + dp[i+1][1-turn], dp[i+1][turn] ) Using memoization, we ensure that overlapping subproblems are efficiently reused. ⚙️ Complexity: ⏱ Time: O(n) 💾 Space: O(n) (can be optimized to O(1)) ✨ Key Insight: What makes this problem elegant is how a simple alternation pattern translates into a strategic inclusion/exclusion decision — showing the beauty of dynamic programming in managing state transitions efficiently. #LeetCodeMediumHardEdition #100DaysChallenge #DynamicProgramming #Recursion #ProblemSolving #Java #LeetCode #CodingChallenge

🚀 Day-65 of #100DaysOfCodeChallenge Today’s problem: 2011. Final Value of Variable After Performing Operations (LeetCode - Easy) This problem was a great reminder that even simple-looking tasks can strengthen our logical reasoning and problem-solving mindset. 🧩 Problem Summary: We’re given a list of operations (++X, X++, --X, X--) that either increment or decrement a variable X. The variable starts at 0, and we need to find its final value after performing all operations on LeetCode. It sounds straightforward — but the beauty of such problems lies in their simplicity. They teach us to: Pay attention to small details. Write clean, efficient logic. Think systematically about every operation in a sequence. 💡 My Approach: I used a simple Java loop to traverse through each operation: If the operation contains '+', increment the value. If it contains '-', decrement it. It’s a one-pass solution — both time and space efficient. ⚙️ Language: Java ⚡ Runtime: 1 ms (Beats 74.53%) 💾 Memory: 42.28 MB (Beats 99.47%) ✅ Status: Accepted (259 / 259 test cases passed) Every solved problem adds one more brick to the wall of consistency and confidence. It’s not just about solving — it’s about understanding, improving, and building the habit of learning every single day. 🔹 Small progress, done daily, creates massive results over time. Onward to Day 66 💪 #100DaysOfCode #LeetCode #CodingJourney #Java #ProblemSolving #TechLearning #DeveloperLife #Consistency #LearningEveryday #ChallengeYourself