Solving LeetCode 396 with O(n) Time Complexity

Day 38 Today I solved: Majority Element (LeetCode 169) 💡 Problem: Given an array, find the element that appears more than [n/2] times. You can assume that the majority element always exists. 💡 My Approach: I used a HashMap (frequency counting) approach: 1️⃣ Traverse the array and store frequency of each element 2️⃣ Calculate n/2 3️⃣ Iterate through the map 👉 Return the element whose frequency is greater than n/2 💡 Key Insight: The majority element will always dominate the count (> n/2), so once counted, it’s easy to identify ✅ ⚡ Complexity: Time: O(n) Space: O(n) #LeetCode #DSA #CodingJourney #LearnInPublic

🚀 LeetCode 438 — Find All Anagrams in a String | Solved ✅ Solved a Medium-level problem using Sliding Window + Hashing. 🔍 Approach: Maintained a dynamic window and tracked character frequencies to efficiently detect anagrams without re-sorting. ⚡ Complexity: Time: O(n) Space: O(1) (optimized using fixed-size array) 💡 Key Learning: Optimizing from hashmap → array significantly improves performance in character-based problems. Step by step, getting better at pattern recognition 🧠 #LeetCode #DSA #SlidingWindow #CodingJourney #ProblemSolving

🚀 Day 25/100 — LeetCode Challenge Today's problem: Find Minimum in Rotated Sorted Array 🧠 Concept: Binary Search on Rotated Array 💡 Key Idea: The minimum element represents the pivot point of rotation. By comparing mid with the rightmost element, we can efficiently narrow down the search space. ⚡ Time Complexity: O(log n) 📂 Solutions Repository https://lnkd.in/gkFh2mPZ Continuing to explore how binary search can be adapted for different problem variations. #100DaysOfLeetCode #DSA #LeetCode #CodingChallenge #SoftwareEngineering

Day 70/100 LeetCode Challenge 🚀 Solved Search in Rotated Sorted Array II using modified binary search. 🔍 Key Learnings: Handling duplicates in rotated arrays is tricky When nums[low] == nums[mid] == nums[high], shrink search space Maintain O(log n) average, but worst case O(n) ⚡ Result: Runtime: 0 ms (Beats 100%) All test cases passed ✅ Consistency > Motivation 💯 #Day70 #LeetCode #BinarySearch #DSA #CodingJourney

📅 Day 270 – Bit Difference Calculation (Apr 3, 2026) ⚙️🔢 🔹 Problems Solved Today: 1️⃣ LeetCode 2220 – Minimum Bit Flips to Convert Number ⚙️ • Computed XOR of start and goal to identify differing bits. • Counted number of set bits in the result. • Each set bit represents one flip needed. • Used built-in bit count or Brian Kernighan’s algorithm. • Reinforced XOR + bit counting technique. 💡 Key Takeaways from Day 270: • XOR directly highlights differences between numbers. • Counting set bits is a fundamental bit manipulation skill. • Many transformation problems reduce to bit-level operations. 🔥 Consistency Update: ✅ Day 270 streak maintained. 🎯 1007 problems solved so far. #LeetCode #DSA #ProblemSolving #BitManipulation #Math #Consistency #Day270 🚀

#100DaysLeetCode Day 47 ✅ Solved LeetCode 141: Linked List Cycle A linked list is given, and we need to check whether it contains a cycle. In other words: does the path end… or does it keep dragging us in circles forever? 🚀 Approach Used Floyd’s Cycle Detection Algorithm: Keep two pointers: slow moves one step at a time fast moves two steps at a time If there is no cycle, fast eventually reaches NULL If there is a cycle, slow and fast will meet at some point One pointer walks. One pointer runs. If they crash into each other, the list has been looping all along. ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) #100DaysLeetCode #Day47 #LeetCode141 #cpp #dsa #linkedlist #twopointers #problemSolving #codingJourney

🚀 Day 25 of #100DaysOfCode Solved Find Minimum in Rotated Sorted Array 🔍 Revisited this classic problem and strengthened my understanding of binary search on rotated arrays. The key idea is to detect which part of the array is sorted and narrow down the search space to find the pivot (minimum element). 🔍 Key Takeaways: If the array is already sorted → first element is the answer Compare mid with neighbors to detect the pivot Use sorted half logic to eliminate unnecessary search space Achieved optimal O(log n) time complexity 💻 Clean logic + edge case handling = efficient solution Consistency is the real game 💯 #DSAwithEdSlash #LeetCode #BinarySearch #100DaysOfCode

📅 Day 298 – Mathematical Transformation & Prefix Optimization (May 1, 2026) 🔄📊 🔹 Problems Solved Today: 1️⃣ LeetCode 396 – Rotate Function 🎯 • Observed relation between consecutive rotations. • Derived formula to update result in O(1). • Used total sum to optimize recalculation. • Avoided brute force rotation simulation. • Iterated once to compute maximum value. 💡 Key Takeaways from Day 298: • Mathematical observation can reduce complexity drastically. • Reusing previous computation avoids redundant work. • Patterns in rotations help derive efficient formulas. 🔥 Consistency Update: ✅ Day 298 streak maintained. 🎯 1036 problems solved so far. #LeetCode #DSA #ProblemSolving #Math #PrefixSum #Consistency #Day298 🚀

🚀 Day 66 of #100DaysOfCode Today, I solved LeetCode 73 – Set Matrix Zeroes, a classic problem that tests in-place matrix manipulation and optimization techniques. 💡 Problem Overview: Given a matrix, if any cell contains 0, its entire row and column must be set to 0. The challenge is to perform this efficiently without using excessive extra space. 🧠 Approach: To solve this optimally, I focused on: ✔️ Using the first row and first column as markers ✔️ Tracking whether the first row/column initially contained zero ✔️ Updating the matrix in-place based on these markers This avoids using additional space and achieves optimal performance. ⚡ Key Takeaways: In-place algorithms help reduce space complexity Using matrix itself as storage is a powerful optimization trick Handling edge cases (first row/column) is critical 📊 Complexity Analysis: Time Complexity: O(n × m) Space Complexity: O(1) Improving problem-solving skills one day at a time 🚀 #LeetCode #100DaysOfCode #DSA #Matrix #ProblemSolving #CodingJourney #SoftwareDevelopment #InterviewPrep

#100DaysLeetCode Day 42 ✅ Solved LeetCode 633: Sum of Square Numbers today. Given a number c, the goal is to check whether it can be written as: a² + b² = c 🚀 Approach Start with a = 0 Keep increasing a while a² <= c For every a, calculate the remaining value: c - a² Check if the remaining value is a perfect square If yes, then we found two numbers whose squares add up to c Otherwise keep moving forward Simple idea: one number climbs the staircase, the other waits hidden in the shadows until the math reveals it. Time Complexity: O(√c) Because we only try values of a from 0 to √c. Space Complexity : O(1) No extra array, no extra data structure. Just a few variables doing all the heavy lifting. Tiny problem. Elegant trick. The kind of question that looks harmless, then quietly tests whether your fundamentals are actually real. #100DaysLeetCode #Day42 #LeetCode633 #cpp #dsa #problemSolving #codingJourney #binarysearch #math