Remove Element Challenge: In-Place Solution

🚀 Day 6 of #100DaysOfCode Solved Find the Duplicate Number on LeetCode using an in-place cyclic sort approach 🔁 🧠 Key insight: When numbers are in the range 1…n, each value belongs at index value − 1. If during placement we find that a number is already at its correct position, that number must be the duplicate. ⚙️ Approach: 🔹Iterate through the array 🔹Place each number at its correct index (num − 1) using swaps 🔹If a conflict occurs (same number already present), return it ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (no extra data structures) #100DaysOfCode #LeetCode #DSA #CyclicSort #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 18 of #100DaysOfCode Solved Remove Duplicates from Sorted List II on LeetCode 🔗 🧠 Key insight: When duplicates appear in a sorted linked list, we need to remove all occurrences, not just one. Using a dummy (sentinel) node makes it easier to handle cases where duplicates start at the head. ⚙️ Approach: 🔹Create a dummy node pointing to the head 🔹Traverse the list with two pointers 🔹If a duplicate sequence is found, skip the entire block 🔹Otherwise, move forward normally ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 60/100 – LeetCode Challenge ✅ Problem: #67 Add Binary Difficulty: Easy Language: Java Approach: Reverse Iteration with Carry Time Complexity: O(max(n, m)) Space Complexity: O(max(n, m)) Key Insight: Binary addition follows same rules as decimal: Sum bits + carry Result bit = sum % 2 New carry = sum / 2 Solution Brief: Iterated from rightmost bits of both strings. Tracked carry and built result from right to left using StringBuilder. Reversed final string for correct order. #LeetCode #Day60 #100DaysOfCode #Binary #Java #Algorithm #CodingChallenge #ProblemSolving #AddBinary #EasyProblem #StringManipulation #BitManipulation #DSA

🚀 Day 20 of #100DaysOfCode Solved 54. Spiral Matrix on LeetCode 🌀 🧠 Key insight: Spiral traversal is all about maintaining boundaries. By shrinking the top, bottom, left, and right limits after each pass, we can cover every element exactly once. ⚙️ Approach: 🔹Maintain four pointers: top, down, left, right 🔹Traverse: 🔹Left → Right (top row) 🔹Top → Bottom (right column) 🔹Right → Left (bottom row) 🔹Bottom → Top (left column) 🔹Update boundaries after each traversal ⏱️ Time Complexity: O(m × n) 📦 Space Complexity: O(1) (excluding output list) #100DaysOfCode #LeetCode #DSA #Matrix #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 27/100 – LeetCode Challenge 🚀 Problem: Sort List Approach: Applied Merge Sort on the linked list Used slow and fast pointers to find the middle Recursively sorted both halves Merged the sorted halves Time Complexity: O(n log n) Space Complexity: O(log n) (recursion stack) Key takeaway: Merge sort is the most efficient sorting technique for linked lists, as it avoids random-access operations required by other algorithms. #LeetCode #100DaysOfCode #DSA #Java #LinkedList #MergeSort #ProblemSolving

🚀 Day 76 of #100DaysOfCode Today I worked on a clean and elegant array manipulation problem — LeetCode: Rotate Array ✅ 📌 Problem Summary Given an array, rotate it to the right by k steps in-place, without using extra space. 🧠 Approach Used: Reverse Technique Instead of shifting elements one by one, I used a smart 3-step reverse strategy: Reverse the entire array Reverse the first k elements Reverse the remaining n − k elements This achieves the rotation efficiently and keeps the solution simple. ⚙️ Complexity ⏱ Time: O(n) 💾 Space: O(1) (in-place) 🔥 Key Learnings Sometimes the best solution is about reordering operations, not adding complexity In-place algorithms are powerful when space constraints matter Reverse logic is a recurring and underrated pattern in array problems Onward to Day 77 🚀 Consistency > Motivation 💪 #100DaysOfCode #LeetCode #Java #DSA #Arrays #ProblemSolving #CodingJourney

Day #42/100 Days of Code🔥 Solved #LeetCode 61: 𝐑𝐨𝐭𝐚𝐭𝐞 𝐋𝐢𝐬𝐭 𝐭𝐨𝐝𝐚𝐲  Focused on understanding linked list manipulation, edge cases, and optimizing the approach to 𝐎(𝐧) 𝐭𝐢𝐦𝐞 & 𝐎(1) 𝐬𝐩𝐚𝐜𝐞. Key takeaways: Always handle 𝐤 % 𝐥𝐞𝐧𝐠𝐭𝐡 in rotation problems Turning the list into a circular linked list simplifies logic #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #Consistency #LearningJourney

🚀 Day 14 of #100DaysOfCode Solved Merge Two Sorted Lists on LeetCode 🔗🔀 🧠 Key insight: Using a dummy (sentinel) node simplifies pointer handling and avoids edge cases when building the merged list. ⚙️ Approach: 🔹Initialize a dummy node to act as the start of the merged list 🔹Compare nodes from both lists one by one 🔹Attach the smaller node and move the pointer forward 🔹Append remaining nodes once one list is exhausted ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 3 of Daily DSA 🚀 Solved LeetCode 1: Two Sum Approach: Used a HashMap to store numbers with their indices. For each element, checked if the complement (target - current) already exists. Complexity: • Time: O(n) • Space: O(n) Performance: Runtime: 2 ms (Beats 99.15%) Memory: 47.34 MB Focusing on writing clean and efficient solutions before over-optimizing. Consistency > Intensity 💯 #DSA #LeetCode #Java #ProblemSolving #Consistency

🚀 Day 70 of #100DaysOfLeetCode Problem: 4Sum Difficulty: Medium Key Concept Learned: Sorting + Two Pointers + Duplicate Handling Today’s challenge was about finding all unique quadruplets in an array whose sum equals a given target. The optimized approach involved sorting the array, fixing two indices, and then using the two-pointer technique to efficiently search for valid combinations. Key takeaways: Sorting enables efficient pointer movement and duplicate detection Skipping duplicates at every level (i, j, left, right) is crucial to ensure uniqueness Two-pointer technique reduces the brute force complexity significantly Careful pointer handling avoids repeated results and improves performance A great problem for strengthening array manipulation, pointer logic, and edge-case handling. On to Day 71! 💪🔥 #LeetCode #DSA #Java #TwoPointers #Arrays #ProblemSolving #CodingJourney #100DaysOfCode