Solved 14. Longest Common Prefix with LeetCode #100DaysOfCode

Day 6 of #100DaysOfLeetCode Problem: 3. Longest Substring Without Repeating Characters Category: Strings / Sliding Window Today’s problem was about finding the length of the longest substring without any repeating characters — a classic question that tests understanding of window management and string traversal. 🧠 Key Learnings: Used a sliding window technique to dynamically adjust the current substring when a duplicate character was found. Tracked characters in the current substring and shifted the start position upon repetition. Focused on maintaining both time efficiency and substring accuracy by managing indices smartly. 🎯 Takeaway: Sliding window techniques are powerful for handling substring and subarray problems efficiently — balance between expanding and shrinking the window is the key. #LeetCode #100DaysOfCode #ProblemSolving #CodingJourney #Strings #SlidingWindow #Python #AIEngineer #Consistency

Day 12 of #100DaysOfLeetCode Problem: 26. Remove Duplicates from Sorted Array Category: Arrays / Two Pointers Today’s problem was about removing duplicates from a sorted array in-place, ensuring each element appears only once and returning the new length. This problem was a great way to strengthen my understanding of two-pointer traversal and array updates without extra space. 🧠 Key Learnings: Used one pointer to track the position of the last unique element while iterating through the array. When a new unique value was found, it was placed at the next available position. Improved confidence in working with in-place modification techniques. Learned to think efficiently — no need for extra arrays or memory allocation. 🎯 Takeaway: Understanding how to manipulate arrays in-place is key to writing space-optimized and interview-ready code. #LeetCode #100DaysOfCode #ProblemSolving #CodingJourney #Arrays #TwoPointers #Python #AIEngineer #Consistency

Day 9 of #30DaysOfCode | LeetCode Problem 3. Longest Substring Without Repeating Characters Implemented an optimized Sliding Window approach to find the length of the longest substring without repeating characters. By maintaining a dynamic window using two pointers and a hash set, the algorithm achieves O(n) time complexity while efficiently managing unique character sequences. #Python #LeetCode #CodingChallenge #ProblemSolving #DataStructures #Algorithms #DSA

Day 4 of #100DaysOfLeetCode Problem: 219. Contains Duplicate II Category: Arrays / HashMap / Sliding Window Today’s challenge was about checking if an array contains duplicate elements within a specific index distance k. I solved this using a hashmap to track the most recent index of each element while iterating through the array. 🧠 Key Learnings: 1) Used a hashmap to perform quick lookups and avoid nested loops. 2) Checked if a duplicate element appeared again within k indices. 3) Optimized the approach to run in O(n) time and O(n) space. 🎯 Takeaway: Tracking element indices in a hashmap helps handle proximity-based duplicate problems efficiently. Staying consistent and learning something new every day! 💪 #LeetCode #100DaysOfCode #ProblemSolving #CodingJourney #HashMap #SlidingWindow #Python #AIEngineer #Consistency

🚀 Day 32 DSA Challenge – Problem #14: Longest Common Prefix Finding unity in diversity — even among strings 😄✨ 🎯 Problem Statement: Given an array of strings, return the longest common prefix shared among them. If there’s no common prefix, return an empty string "". 🧠 How I Solved It: Started by assuming the first string as the initial prefix. Iterated through each subsequent string, trimming the prefix until it matched the start of the current word. Used Python’s startswith() to efficiently check prefix matches. Stopped early if the prefix became empty — ensuring optimal performance. ⚙️ Performance Stats: ⏱ Runtime: 0ms (beats 100%) 💾 Memory: 12.52MB (beats 27.15%) ✅ Testcases Passed: 126 / 126 💡 What I Learned: This problem reinforced the value of incremental reduction and early termination in string algorithms. Sometimes, solving efficiently means knowing when to stop ⏳💬 #LeetCode #Problem14 #LongestCommonPrefix #StringManipulation #Python #AlgorithmDesign #100DaysOfCode #DSA 

LeetCode 90-Day Challenge – Day 78 Problem: Find Peak Element Difficulty: Medium Problem: We need to find a peak element in an array, an element that is strictly greater than its neighbors. If multiple peaks exist, returning the index of any one of them is valid. Conceptually, nums[-1] and nums[n] are considered negative infinity, so edge elements can also be peaks. The challenge is to achieve this in O(log n) time. Solution approach: This problem can be efficiently solved using binary search. At each step, we check the middle element and compare it with its right neighbor. If the middle element is smaller than the next one, it means the peak lies on the right half; otherwise, the peak is on the left half. By continuously narrowing down the range, we find a peak element in logarithmic time. #LeetCode #90DaysOfCode #Python #LinkedInChallenge #CodingJourney #ProblemSolving #LeetCodeChallenge

🚀 LeetCode #3461 – Check If Digits Are Equal in String After Operations I Today’s challenge was all about pattern transformation using modulo logic! 🔢 We repeatedly take each pair of consecutive digits, find their sum modulo 10, and form a new sequence, until only two digits remain. If both digits are the same, return True; otherwise, return False. ✨ It’s a neat way to test your understanding of string manipulation, loops, and modular arithmetic. 📊 Complexity: Time – O(n²) Space – O(n) Sometimes even the simplest looking problems hide cool math logic behind them 😄 Have you tried this one yet? What was your approach? 👇 #LeetCode #Python #CodingChallenge #ProblemSolving #LogicBuilding #100DaysOfCode #DSA

🚀 Day 73 of #100DaysOfDSA 🚀 LeetCode 240 — Search a 2D Matrix II 🔹 Problem: Find whether a target exists in a sorted 2D matrix (rows and columns are sorted). 🔹 Approach: Staircase Search (Top-Right Start) Start from the top-right corner: If the value is greater than target → move left If smaller → move down Continue until found or out of bounds ✨ Key Insight: Sorted rows and columns allow a greedy elimination of one row or column each step — making it efficient! #LeetCode #DSA #Matrix #BinarySearch #Python #ProblemSolving #100DaysOfCode #CodingJourney 🚀