Remove Element in-place array manipulation with Python

🚀 Day 85 of #100DaysOfLeetCode 🔍 Problem Solved: Ransom Note (LeetCode 383) Today’s problem was all about efficiently checking whether one string can be constructed from another — a classic hashing / frequency counting concept. ⚡ What I Learned: - Importance of frequency maps (hash tables) - Writing optimized solutions over naive approaches - How built-in methods can simplify logic but may impact performance 📊 Performance: ✅ Runtime: 0 ms (Beats 100%) ✅ Memory: Efficient usage 🔥 Takeaway: Small optimizations and choosing the right data structure can make a huge difference, even in easy problem #Day85 #LeetCode #CodingJourney #Python #DataStructures #ProblemSolving #100DaysOfCode

🚀 LeetCode Progress Update – Problem Solved! ✅ Problem: Remove Trailing Zeros From a String 💡 Approach: Used reverse traversal to find the first non-zero digit and sliced the string accordingly. 🔍 Key Learning: Efficient string manipulation can avoid unnecessary conversions. Traversing from the end helps solve trailing-based problems quickly. 💻 Code Insight: Instead of removing zeros one by one, I identified the breakpoint and sliced the string — making it optimal and clean. ⏱️ Performance: Runtime: 3 ms ⚡ Beat: 66%+ users Memory: 19.23 MB 📈 Consistency is key — one problem closer to mastery! #LeetCode #CodingJourney #Python #ProblemSolving #DSA #100DaysOfCode

🚀 Day 62 of #100DaysOfCode Solved “Remove Nth Node From End of List” 🔗 💡 Today’s focus: Two Pointer Technique (Fast & Slow pointers) Instead of calculating length, I used an efficient one-pass approach to remove the target node. 🧠 Key Learnings: Dummy node helps handle edge cases (like removing head) Fast pointer moves n steps ahead Then move both pointers until fast reaches the end Slow pointer lands just before the node to delete ⚡ Clean, efficient & optimal solution (O(n) time, O(1) space) Consistency is starting to feel powerful now 💪 #DSA #LeetCode #CodingJourney #LinkedInLearning #100DaysOfCode #Day62 #Python #ProblemSolving

🚀 Day 38 – LeetCode Journey Today’s problem: Gray Code ✔️ Generated sequence using bit manipulation ✔️ Applied formula: "i ^ (i >> 1)" ✔️ Ensured only one bit changes between consecutive numbers 💡 Key Insight: Gray Code is useful in minimizing errors in digital communication, as only one bit changes at a time. Using bitwise operations makes the solution both elegant and efficient. This problem improved my understanding of bit manipulation and binary patterns. Exploring deeper into low-level concepts 🔥💪 #LeetCode #Day38 #BitManipulation #Binary #Python #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 41 – LeetCode Journey Today’s problem: Add Binary ✔️ Implemented binary addition manually ✔️ Handled carry propagation efficiently ✔️ Worked with string manipulation and indexing 💡 Key Insight: Binary addition follows the same rules as decimal addition, but with base 2. Managing the carry correctly is the key to building the final result. This problem improved my understanding of bit-level operations and string handling. Step by step, mastering the fundamentals 🔥💪 #LeetCode #Day41 #Binary #Strings #Python #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Just solved the “Valid Number” problem on LeetCode! This problem looks simple at first glance—but handling edge cases like decimals, signs, and exponents makes it a great test of attention to detail and logical thinking. ✅ Key takeaways: Careful handling of edge cases is crucial Validating input step-by-step can simplify complex parsing problems Writing clean, readable logic beats overcomplicated solutions 💡 Performance: ⚡ Runtime: 3 ms 🧠 Efficient space usage ✅ All test cases passed Problems like this remind me that consistency in practice is what builds strong problem-solving skills. On to the next one! 🔥 #LeetCode #Coding #Python #ProblemSolving #SoftwareEngineering #AIEngineerJourney link of #Solution :- https://lnkd.in/ga9b5pVb

🚀 Day 89 of #100DaysOfCode Solved LeetCode Problem 231 — Power of Two ✅ Cracked this one with a clean and efficient bit manipulation approach. The goal was to check whether a number is a power of two without using loops or recursion. 💡 Key Insight: A power of two has exactly one set bit in its binary form. So, using the trick: 👉 "n > 0 and (n & (n - 1)) == 0" This removes the lowest set bit — if the result is 0, it's a power of two. ⚡ Performance: • Runtime: 0 ms (Beats 100%) • Memory: 19.27 MB 🔍 What I learned: • Bit manipulation is insanely powerful • Writing optimal solutions matters • Small tricks can save big time Consistency > Motivation. On to Day 90 🔥 #LeetCode #DSA #CodingJourney #100DaysChallenge #BitManipulation #Python

LeetCode Progress Update — 18/100 Just solved “415. Add Strings” This problem was a great reminder that sometimes the best solutions come from going back to fundamentals — in this case, simulating manual addition digit by digit instead of relying on built-in conversions. Key Takeaways: Handling numbers as strings strengthens problem-solving skills Mastering carry logic is essential for many algorithmic problems Clean, efficient logic > shortcuts Problem Level: Easy Concepts: String Manipulation, Simulation Every step forward counts. On to the next one! #100DaysOfCode #DSA #LeetCode #ProblemSolving #CodingJourney #Python

Top K Frequent: Bucket Sort Beats Heap with O(n) Time Heap-based solution costs O(n log k). Bucket sort achieves O(n) by leveraging frequency bounds — frequencies can't exceed array length. Index = frequency, value = list of elements with that frequency. Traverse buckets high-to-low, collecting k elements. Bucket Sort Advantage: When value range is bounded and known (frequencies ≤ n), bucket sort beats comparison-based O(n log k). Exploiting constraints transforms complexity. Time: O(n) | Space: O(n) #BucketSort #TopK #FrequencyAnalysis #ComplexityReduction #Python #AlgorithmOptimization #SoftwareEngineering

🚀 Day 33 – LeetCode Journey Today’s problem: Move Zeroes ✔️ Used two-pointer technique for in-place modification ✔️ Moved all non-zero elements forward efficiently ✔️ Maintained relative order of elements 💡 Key Insight: By keeping a pointer for non-zero elements, we can swap values in a single pass and avoid extra space — making the solution both clean and optimal. This problem improved my understanding of array manipulation and in-place algorithms. Consistency is building confidence day by day 💪🔥 #LeetCode #Day33 #Arrays #TwoPointers #Python #ProblemSolving #CodingJourney #100DaysOfCode