Discovering Rust: A Philosophy of Programming

📅 Day 49 of #100DaysOfCode Problem: 3228. Maximum Number of Operations to Move Ones to the End (LeetCode) Approach: 1️⃣ Keep a counter of how many '1's you’ve seen as you go through the string. 2️⃣ When you see a '0' that’s right before a '1' or at the end, you can “move” all those previous '1's past this '0' — so add that counter to the result. 3️⃣ Every time you hit a '1', increment the counter; for '0', check the condition and reset logic as needed. 4️⃣ Return the total operations counted — it’s just one pass through the string = O(n) time, O(1) space. #100DaysOfCode #LeetCode #DSA #ProblemSolving #Cplusplus #Algorithms #Greedy #CodingChallenge #Programming #DeveloperLife #CodeNewbie #DailyDSA #SoftwareEngineering #KeepLearning #TechCommunity #GrowthMindset #Motivation

🚀 𝐌𝐚𝐬𝐭𝐞𝐫 𝐎𝐛𝐣𝐞𝐜𝐭-𝐎𝐫𝐢𝐞𝐧𝐭𝐞𝐝 𝐏𝐫𝐨𝐠𝐫𝐚𝐦𝐦𝐢𝐧𝐠 (𝐎𝐎𝐏𝐬) 𝐢𝐧 𝐣𝐮𝐬𝐭 𝟏𝟓 𝐝𝐚𝐲𝐬! Whether you're preparing for tech interviews or strengthening your programming foundation, understanding OOPs concepts — Encapsulation, Abstraction, Inheritance, and Polymorphism — is a must. With AlgoTutor’s structured notes & practice plan, you can go from basics to mastery — one concept at a time 💪 📘 𝐋𝐞𝐚𝐫𝐧 | 💡 𝐏𝐫𝐚𝐜𝐭𝐢𝐜𝐞 | 🚀 𝐌𝐚𝐬𝐭𝐞𝐫 #AlgoTutor #Programming #OOPs #DSA #Coding #TechSkills #Learning #SystemDesign #EdTech

🔥 Day 70 of #100DaysOfCode 🔥 💡 Problem: Number of Steps to Reduce a Number to Zero – LeetCode ✨ Approach: A simple while loop logic — divide the number by 2 if it’s even, else subtract 1. Repeated until it hits zero. Clean, elegant, and lightning fast! ⚡ 📊 Complexity Analysis: Time Complexity: O(log n) – each division by 2 halves the number Space Complexity: O(1) – constant space ✅ Runtime: 0 ms (Beats 100%🔥) ✅ Memory: 41.95 MB 🚀 Key Takeaway: Sometimes, brilliance lies in simplicity — clear logic, powerful performance! #LeetCode #100DaysOfCode #ProblemSolving #CodingChallenge #Programming #DSA #LogicBuilding #Efficiency #CodeDaily #DeveloperJourney

On Day 301 of the coding journey, I'm sharing my solution to LeetCode Problem 658, "Find K Closest Elements." This problem is an excellent example of using Binary Search to quickly establish a search range, which is then refined by the Two-Pointer technique. My video provides a clear walkthrough of this efficient $O(\log N + K)$ approach, a valuable skill for technical interviews and a great way to think about optimizing array search problems. #DataStructures #Algorithms #LeetCode #CodingInterview #Programming #BinarySearch #TwoPointers

On Day 285 of my #300daysofcode challenge, I'm sharing my solution to LeetCode Problem 2048, "Next Greater Numerically Balanced Number." This problem requires a focused iterative search and a helper function to verify the unique "balanced" property (where each digit $d$ appears $d$ times). My video provides a clear walkthrough of this logical method, a valuable skill for any developer and a great way to practice writing clean, reusable validation code. #DataStructures #Algorithms #LeetCode #CodingInterview #Programming #ProblemSolving #300daysofcode

📌 Master Technology: Key takeaways from this post. In 2025, development is no longer limited to local environments. Thanks to Devtools as a Service… Read the full article: https://lnkd.in/dFzg7GBP Read the full article: https://lnkd.in/dvMZFf2E #WebDevelopment #Programming #TechTips #Developer #Coding

On Day 304 of the coding journey, I'm sharing my solution to LeetCode Problem 852, "Peak Index in a Mountain Array." This problem is a great test of recognizing and leveraging the monotonic property inherent in the array's structure. My video provides a clear walkthrough of the Binary Search adaptation, which is a valuable skill for technical interviews and essential for solving problems in $O(\log N)$ time. #DataStructures #Algorithms #LeetCode #CodingInterview #Programming #BinarySearch #Optimization

🚀 Day 62 of LeetCode150DaysChallenge Problem: Rotate a Linked List Right #DSA #LeetCode #CodingChallenge #LinkedList #C++ #Programming #150DaysOfCode #LearnDSA #SDEJourney Topic: Linked List Given the head of a linked list, we need to rotate it to the right by k places. Similar to rotate an array by k elements For example: Input: 1 → 2 → 3 → 4 → 5, k = 2 Output: 4 → 5 → 1 → 2 → 3 🔍 Intuition: Rotating a linked list is basically bringing the last k nodes to the front. To do that efficiently, we need to understand: The length (n) of the list. The point of split (at position n - k). Reconnect the parts properly. 🧠 Step-by-Step Approach: Find the length of the list. Compute k = k % n to handle rotations larger than the list size. Break the list at the (n - k)th node. Reverse both halves for easier concatenation. Join them, and finally reverse the whole list back to get the rotated version. This clever use of the reverse operation avoids complex pointer adjustments! ⏱️ Time Complexity: O(n) – one traversal to find length, and a few reversals. 💾 Space Complexity: O(1) – in-place manipulation. 💡 Key Takeaway: This problem beautifully shows how reversing sublists can simplify complex pointer operations. Mastering linked list manipulations like these builds strong problem-solving intuition! 💪

🔍 Debugging can be frustrating, but it’s also one of the best teachers in programming. Every bug you fix teaches you how systems really work — not just how they should. My go-to process: 1️⃣ Reproduce the issue 2️⃣ Isolate the cause 3️⃣ Read the logs carefully 4️⃣ Fix, test, document What’s your favorite debugging trick? 💭 #Coding #DeveloperLife #ProblemSolving

Day 4 of the #Code2Grow Challenge Today’s challenge was to write a C program that calculates a student’s percentage and grade based on marks and attendance. It took some thinking to get all the nested if–else conditions right, especially the part with attendance and grace marks, but I finally cracked it. It was a great exercise in logical thinking and understanding how real-world conditions can be turned into code. Here’s a short video where I explain my logic. #Code2Grow #SRMAPCoding #Day4Challenge #CProgramming #LogicBuilding #CodingJourney #ProblemSolving #LearnByDoing Dr. Yasir Afaq