Derive Rust Rules with Space Time and Coordinates

rust 1.95.0 released ✅ cfg_select! is now built-in: Say goodbye to dependency bloat. This new macro provides a native way to handle conditional compilation with a clean, match-like syntax. It’s essentially a built-in version of the popular cfg-if crate. ✅ if let guards in matches: We finally have more power in match arms! You can now use if let patterns as guards, allowing you to bind variables and check conditions simultaneously within a single match arm. No more nested "if" soup. ✅ Fixing the "Range" Mistake: For years, the fact that Range didn't implement Copy was considered a core design regret. 1.95.0 introduces core::range::RangeInclusive, a major step toward a modern, copyable range system that will eventually become the default in Edition 2027. https://lnkd.in/ddtBn_R8 #RustLang #Programming #SoftwareEngineering #OpenSource #CodingUpdate #Rust195

Rust's borrow checker isn't the hard part. The hard part is unlearning every shortcut you got away with in other languages. No more "I'll just clone it." No more "this pointer is definitely still valid." No more silent data races. Rust doesn't trust you. After years of C++ bugs — honestly, fair. #Rust #Programming #SoftwareEngineering #SystemsProgramming #DevLife

Every great programmer starts with the basics. Diving deep into Singly Linked Lists — understanding how nodes connect, how memory is managed dynamically, and how operations like insertion and deletion actually work behind the scenes. It’s not just about code, it’s about building logic. awareness of hardware-level behavior like cache alignment and data locality. #Multithreading #Concurrency #Programming #SoftwareEngineering #BackendDevelopment #CPP #CppProgramming #ModernCPP #Multithreading #Concurrency #ParallelProgramming #STL #SystemProgramming #SoftwareEngineering #CodingLife #DSA #Coding #LinkedList #TechLearning #ProgrammerMindset

Invert Binary tree Approach: check if node is null or not if null return null swap left node and right ( not value complete node ) go to left (recursively) go to right (recursively) return node TC: O(N) SC:O(N)- recursive stack #DSA #programming #coding #problemSolving

🚀 Day 93 of #100DaysOfCode Challenge Today’s problem was all about identifying “Beautiful Strings” 🔢✨ A numeric string is called beautiful if: ✔️ It can be split into a sequence of increasing numbers ✔️ Each number is exactly +1 from the previous ✔️ No leading zeros allowed 💡 What I Learned Today: How to break strings into valid sequences Handling large numbers using long long Importance of string comparison vs integer operations Edge cases like: Leading zeros ❌ Single digit strings ❌ Invalid increments ❌ 🧠 Approach: Try all possible starting numbers Generate the sequence dynamically (x, x+1, x+2…) Match the built string with the original If matched → ✅ YES x Else → ❌ NO 💻 Example: 👉 Input: 91011 👉 Output: YES 9 ⚡ Key Takeaway: Sometimes brute force with smart validation is the best approach! 📅 Consistency is the real game changer. On to Day 94 💪 #Coding #Programming #CProgramming #DataStructures #ProblemSolving #100DaysOfCode #DeveloperJourney #LearningEveryday

LCA - Lowest Common Ancestor of binary tree Approach: If current node is null, return null If current node is p or q, return current node Recursively search left and right If both sides return non-null → current node is LCA Otherwise → return the non-null side TC = O(n) SC = O(h) - depends on tree height (recursion stack) O(logn) - balanced tree O(n) - skewed tree #dsa #leetcode #consistency #coding #programming

Subtree of another tree Approach: If subRoot is null - return true If root is null - return false If values match - check isIdentical(root, subRoot) If identical - return true Else - check left and right: isSubtree(root.left, subRoot) isSubtree(root.right, subRoot) Return true if found anywhere, otherwise false TC: O(N*M) -- n =no. of nodes in root, m = no. nodes in subRoot (worst case) SC: O(h) -- height of tree ( O(N) -- skewed tree || O(logN) -- balanced tree) #coding #programming #DSA #Consistency #Leetcode #CodingJourney

Solved a Gray Code problem in C++ today. The task was to generate bit patterns from 0 to 2^n - 1 such that every consecutive pattern differs by only one bit, while always starting from 0. I used the Gray code formula: gray = i ^ (i >> 1) This makes the solution clean and efficient, and guarantees that adjacent codes differ by exactly one bit. Example for n = 2: 00 -> 01 -> 11 -> 10 What I like about this problem is how a simple bit manipulation formula can solve what looks like a complex sequence-generation challenge. Concepts practiced: Bit Manipulation Binary Representation Pattern Generation C++ Problem Solving #cpp #coding #programming #datastructures #algorithms #problemsolving #bitmanipulation #leetcode #geekforgeeks

Solved LeetCode 151: Reverse Words in a String Today, I worked on reversing the order of words in a string while handling extra spaces efficiently. I learned how to: Remove leading, trailing, and multiple spaces Extract words correctly Reverse their order using clean logic and STL I implemented an optimized solution with: Time Complexity: O(n) Space Complexity: O(n) This problem helped me improve my understanding of string manipulation, edge case handling, and clean coding practices https://lnkd.in/getBKFiE GitHub Repo: https://lnkd.in/gg4daDpn #day18 #DSA #Cpp #LeetCode #Coding #Programming #Learning #ProblemSolving #Strings

Binary tree right side view Approach: - Keep a level variable - If level == result.size(), add current node (first node of that level) - Traverse right first, then left Time: O(n) Space: O(h) (worst: O(n), best: O(log n)) #Algorithms #DSA #LeetCode #coding #Programming