Remove Element in LeetCode Array

Headline: 🚀 From Confused to Confident: Solving LeetCode 2906! Body: Ever stared at a problem statement and thought, "Where do I even start?" That was me today with LeetCode 2906: Construct Product Matrix. The task seemed simple: For every cell in a matrix, calculate the product of all other elements. But the catch? Doing it efficiently without hitting a Time Limit Exceeded (TLE) error. 💡 The Breakthrough: I realized this is essentially a 2D version of the classic "Product of Array Except Self." Instead of brute-forcing it (which would be O(N²)), I used the Prefix & Suffix Product approach. 🎓 How I visualized it (The Story): Imagine a classroom photo where every student needs to calculate the "score" of everyone else except themselves. 1️⃣ Forward Pass: Calculate the product of everyone standing to your left. 2️⃣ Backward Pass: Calculate the product of everyone standing to your right. 3️⃣ Multiply: Left × Right = Your Answer! By treating the 2D matrix as a continuous stream, I optimized the solution to: ✅ Time Complexity: O(m × n) ✅ Space Complexity: O(1) (excluding output matrix) 🔑 Key Takeaway: Don't rush to code. Spend time visualizing the data flow. Sometimes, breaking a 2D problem into a 1D logic stream makes all the difference. Has anyone else tackled this problem? What was your approach? Let's discuss in the comments! 👇 #LeetCode #DSA #Coding #ProblemSolving #SoftwareEngineering #LearningJourney #Python #Algorithms #TechCommunity

🔥 Day 8/100 of My LeetCode Challenge — 0 ms (100%) 🔥 Hook: Most people fail this “easy” problem because they ignore one tiny detail — can you spot it? 💭 Problem: Add 1 to a number represented as an array of digits. Sounds simple, right? But here’s the catch 👇 👉 What if the last digit is 9? 👉 What if ALL digits are 9? Example: [9,9,9] → [1,0,0,0] 💥 🧠 What I learned today: Always think about edge cases Simple problems can hide tricky logic Traversing backwards can simplify carry problems ⚡ My Approach: Start from the end If digit < 9 → add 1 and return If digit == 9 → make it 0 and continue If all become 0 → add 1 at front 💻 Result: ✅ Runtime: 0 ms (100%) ✅ Clean & optimized solution 📈 Consistency Check Day 8 complete. No excuses. No breaks. Small wins daily → Big results later 🚀 #LeetCode #Day8 #CodingJourney #Python #ProblemSolving #100DaysOfCode

Day 64 of LeetCode Grind ⚡🔥 Two classic Medium problems today — both are interview must-knows and each teaches a pattern you'll reuse constantly. 1️⃣ Top K Frequent Elements (347) <<< return [x for x, _ in Counter(nums).most_common(k)] >>> * Counter.most_common(k) uses a heap internally to find top-k in O(n log k) — better than sorting which is O(n log n). Want true O(n)? Use Bucket Sort: > Create n+1 buckets indexed by frequency. > Elements with frequency f go into bucket[f]. > Scan buckets from highest frequency down, collect until you have k elements. > Since frequency can't exceed n, bucket index is bounded → pure O(n)! 2️⃣ Product of Array Except Self (238) > No division allowed. O(n). O(1) extra space. Classic prefix + suffix product pattern: * nums = [1, 2, 3, 4] * prefix = [1, 1, 2, 6] ← everything LEFT of i * suffix = [24, 12, 4, 1] ← everything RIGHT of i * result = [24, 12, 8, 6] ← multiply both * Two passes, one output array. The trick is reusing the result array itself — store prefix in the first pass, multiply suffix in-place on the second pass. ✨ Reflection: Both problems follow the same meta-pattern: precompute something so each query is O(1). Whether it's frequency counts or prefix products, the "precompute → combine" mindset is fundamental to efficient algorithm design. #LeetCode #Day64 #Python #PrefixProduct #BucketSort #HashMap #Arrays #Medium #100DaysOfCode #DSA

STOP THE "STORAGE BOX" LIE! 📦🚫 Your Python variables aren't physical boxes; they are frantic goblins holding onto memory tethers. It’s time to see the strings and stop the bugs before you smash another digital potion! 🧵👹. Ever change one variable and accidentally break an entirely different function?. Your code isn’t haunted, it’s just a tug-of-war between reference tethers!. We just dropped "The Alchemist’s Memory" masterclass on YouTube to help you visually decode: • The Assignment Tether: How goblins link names to data chests. • Logic Scissors: How to properly "snip" references without crashing your system. • The Spectral Hand: How the Garbage Collector sweeps away "abandoned" data. 🔥 Watch the 60-second cinematic breakdown here: https://lnkd.in/d8Jr_NgX Don't just write code, master the architecture. Download the accompanying Code Codex & Chronicles worksheet and start snipping those tethers like a pro! ✂️📜. Keep Building, The Nazli Tech School Team 🎓💻

🚀 Ever wondered how to optimize your code for performance? Let's dive into the concept of algorithm efficiency! 📈 Algorithm efficiency is all about writing code that runs faster and uses less memory. It's crucial for developers as efficient algorithms can make your applications more responsive and reduce server costs. 💰 Here's a step-by-step breakdown: 1️⃣ Analyze the problem and understand the requirements. 2️⃣ Choose the right algorithm to solve the problem. 3️⃣ Implement the algorithm efficiently in your code. Check out this code snippet for calculating Fibonacci numbers efficiently: ```python def fibonacci(n): a, b = 0, 1 for _ in range(n): a, b = b, a + b return a ``` Pro Tip: Remember to test your algorithms with different inputs to ensure they perform well in various scenarios. 🧪 Common mistake to avoid: Neglecting to consider the scalability of your algorithm can lead to performance issues later on. Always think about how your code will behave with larger datasets. 📉 🤔 What are some ways you optimize your code for efficiency? Share your tips below! 💬 🌐 View my full portfolio and more dev resources at tharindunipun.lk #AlgorithmEfficiency #CodeOptimization #DeveloperTips #PerformanceMatters #CodingLife #TechTalk #CodeSnippets #LearnProgramming #ProblemSolving #AlgorithmDesign

Hello dudes and dudettes!! 🚀 Day 13/150 — Solved LeetCode 238: Product of Array Except Self Today’s problem was a perfect mix of simplicity and smart thinking 🤯 At first glance, it feels straightforward: 👉 For each element, multiply all the other elements But the catch? ⚡ You can’t use division ⚡ And you need to do it in O(n) time That’s where things get interesting. 🧠 Initial Thought My first instinct was: “Okay, just loop through the array and multiply everything except the current element.” But that leads to O(n²) time complexity ❌ Too slow for large inputs. 💡 The Breakthrough Moment Instead of recalculating everything again and again, I realized: 👉 Why not reuse previous computations? This leads to a powerful idea: Compute products from the left side Compute products from the right side Combine both 🔥 The Core Idea For every index: 👉 Answer = (product of elements on the left) × (product of elements on the right) No division. No repeated work. Just smart reuse. 📊 Example Input: [1, 2, 3, 4] We mentally break it like this: Left products → builds progressively from the start Right products → builds from the end Then combine both to get: 🎯 Output: [24, 12, 8, 6] 😎 What Made This Problem Interesting? It looks simple, but forces you to think efficiently Teaches how to avoid redundant computation Shows how powerful prefix & suffix patterns are 💡 What I Learned Don’t recompute — reuse Think in terms of patterns, not brute force Sometimes the best solution comes from splitting the problem into two passes 🎯 Key Takeaway “Efficiency isn’t about doing more work faster… it’s about doing less work smarter.” 🔥 Another step forward in the journey — learning to think, not just code. #LeetCode #Algorithms #DataStructures #ProblemSolving #CodingJourney #100DaysOfCode #Python #LearningInPublic

🚀 Day 2 of My LeetCode Journey Today I solved 2 interesting problems that strengthened my understanding of data structures and logic building 👇 🔹 1. Insert Delete GetRandom (O(1)) This problem was all about designing an efficient data structure. I learned how to combine: ✔️ HashMap (for O(1) lookup) ✔️ Array/List (for O(1) random access) 💡 Key takeaway: To achieve constant time complexity, sometimes you need to blend multiple data structures smartly. 🔹 2. Check If All 1's Are at Least Length K Places Away This problem focused on iteration and condition checking. 💡 Key takeaway: Track the position of the last seen 1 and ensure the distance condition is maintained. 🔥 What I’m improving daily: • Problem-solving mindset • Writing optimized code • Thinking in edge cases Consistency > Perfection 💯 #Day2 #LeetCode #Python #CodingJourney #ProblemSolving #100DaysOfCode

🚀 Another LeetCode Problem Solved: Palindrome Number! 🔗 Check out my solution: https://lnkd.in/dwDMqXXn 💡 Problem Overview Given an integer x, determine whether it is a palindrome — meaning it reads the same forward and backward. (LeetCode) Examples: ✔ 121 → Palindrome ❌ -121 → Not a palindrome ❌ 10 → Not a palindrome 🧠 My Approach (Digit Reversal) Instead of converting the number to a string, I used a mathematical approach: Extract digits using % 10 Reverse the number step by step Compare reversed number with original ⚙️ Key Learnings ✔ Strong understanding of number manipulation ✔ Importance of handling edge cases (negative numbers, trailing zeros) (leet-solution.com) ✔ Practicing clean and efficient logic ⏱️ Complexity • Time Complexity: O(log n) • Space Complexity: O(1) 🔥 Why this problem matters Even though it’s an “easy” problem, it builds: Logical thinking Problem-solving fundamentals Confidence for bigger challenges 🚀 Next Goal Move towards optimized approaches and medium-level problems to strengthen DSA skills. #LeetCode #DSA #Python #CodingJourney #ProblemSolving #100DaysOfCode

Hello dudes and dudettes!! 🚀 Day 11/150 — Cracked LeetCode 274: H-Index Today’s problem was less about coding… and more about thinking smart 😄 At first glance, it looks like just another array problem. But once I dug in, I realized it’s actually about measuring impact — not just numbers. 🔍 What’s the Problem About? You’re given an array where each number represents how many times a research paper was cited. Now the twist: 👉 You need to find the H-Index Which basically means: A researcher has an H-Index of h if they have h papers with at least h citations each. 🧠 The “Aha!” Moment Initially, it feels confusing… like, “Where do I even start?” But everything clicks when you: 👉 Sort the array in descending order Suddenly, the problem becomes super clean. Now it’s just about checking: “Do I have 1 paper with ≥1 citation?” “Do I have 2 papers with ≥2 citations?” “Do I have 3 papers with ≥3 citations?” …and so on. The moment this condition fails — boom, you stop. That’s your answer. 📊 Quick Example Input: [3, 0, 6, 1, 5] After sorting: [6, 5, 3, 1, 0] Now checking step-by-step: 1 paper → valid ✅ 2 papers → valid ✅ 3 papers → valid ✅ 4 papers → not valid ❌ 🎯 Final Answer: 3 💡 What Made This Problem Interesting? It’s not about checking everything — it’s about knowing when to stop A simple sort can completely change how you see the problem It teaches you to think in terms of thresholds and consistency, not just values 😎 My Takeaway This problem is a perfect reminder that: “Sometimes the smartest solution isn’t more work… it’s better perspective.” 🔥 One more step forward in the journey. Let’s keep building. #LeetCode #Algorithms #ProblemSolving #CodingJourney #100DaysOfCode #Python #LearningInPublic

Hello dudes and dudettes!! 🚀 Day 17/150 — Solved LeetCode 13: Roman to Integer Today’s problem was a fun twist — not your usual numbers, but ancient ones 😄🏛️ Converting Roman numerals into integers sounds simple… until the rules start playing tricks on you. 🧠 What’s the Problem About? You’re given a Roman numeral string like: 👉 “III”, “IV”, “IX”, “MCMIV” And your job is to convert it into a normal integer. 💡 Initial Thought At first, I assumed: “Just map each symbol and keep adding.” And that works… sometimes. But then comes the twist 👇 🔥 The Catch Roman numerals don’t always follow simple addition. 👉 If a smaller value comes before a bigger value, you subtract instead of adding. Examples: IV = 5 - 1 = 4 IX = 10 - 1 = 9 That’s where things get interesting. 💥 The Breakthrough Moment Instead of overthinking, I simplified the logic to one powerful idea: Look at the next character. If the next value is bigger → subtract current Otherwise → add current That’s it. Just one small check completely solves the problem. ⚙️ How It Works Traverse the string from left to right Compare each character with the next Decide whether to add or subtract Keep updating the total Simple logic. Clean execution. 📊 Why This Problem Is Cool It mixes logic with pattern recognition Shows how a tiny condition can change the whole approach Teaches you to look ahead before deciding 😎 What I Learned Don’t assume rules are always straightforward Edge cases often define the real solution Breaking the problem into small decisions makes everything easier 🎯 Key Takeaway “Sometimes the answer isn’t in the current step… it’s in what comes next.” 🔥 Another step forward — learning to think, not just code. #LeetCode #Algorithms #ProblemSolving #CodingJourney #100DaysOfCode #Python #LearningInPublic