Solving LeetCode 476: Number Complement with Bit Manipulation

💡 LeetCode #88 — Merge Sorted Array Today I solved LeetCode Problem 88: Merge Sorted Array 🧩 Problem Summary: Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. All merging should happen in-place, meaning you can’t use extra space for another array. Key Idea: Use three pointers starting from the end: i → last valid element in nums1 j → last element in nums2 k → last index in nums1 (total length) Compare elements from the back and fill nums1 from the end — this avoids overwriting elements we haven’t processed yet. #LeetCode #Java #DSA #TwoPointers #CodingJourney #ProblemSolving

📌 Day 2/100 – Remove Element (LeetCode 27)  🔹 Problem:  Given an integer array nums and a value val, remove all instances of that value in-place and return the new length of the array. The order of elements can be changed. 🔹 Approach: Used the two-pointer technique to efficiently modify the array in-place. One pointer iterates through the array, while the other tracks the position to overwrite non-val elements. Returned the position of the second pointer as the new length. 🔹 Key Learning: Strengthened understanding of in-place array manipulation. Improved logic building for pointer movement and conditional overwriting. Learned how to minimize extra space usage while maintaining readability and clarity. Another small yet powerful step toward mastering array-based problems! 💻 🔥 #100DaysOfCode #LeetCode #Java #ProblemSolving #TwoPointers #DSA #CodingJourney

🚀 115 days of #200DaysOfCode Problem: 24. Swap Nodes in Pairs Problem Statement: Given the head of a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes, only nodes themselves may be changed. Approach: Used a dummy node and iteratively swapped each adjacent node pair via pointer manipulation, which enabled in-place node swaps without extra space. Logic: Leveraged pointer rewiring to achieve the swaps efficiently with O(1) extra space and O(n) time complexity, cleanly iterating through the list to handle both even and odd-length cases. 👉 Question link 🔗: https://lnkd.in/g6cwvMgz #LeetCode #Java #LinkedList #Pointers #DSA #Coding #Algorithms #InterviewPrep #200DaysOfCode

📌 Day 18/100 - Valid Palindrome (LeetCode 125) 🔹 Problem: Determine whether a string reads the same forward and backward, ignoring case and non-alphanumeric characters. 🔹 Approach: Implemented a two-pointer technique. Skipped all non-alphanumeric characters. Compared characters from both ends in lowercase. Returned true if all matched, otherwise false. 🔹 Key Learning: Two-pointer method keeps logic clean and efficient. Character handling is key when data isn’t uniform. Time complexity: O(n), Space complexity: O(1). Sometimes, solving elegantly is better than solving fast. ✨ #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney

💡 LeetCode #977 — Squares of a Sorted Array Today I solved LeetCode Problem 977: Squares of a Sorted Array ⚡ Problem Summary: Given a sorted array of integers nums, return an array of the squares of each number, also sorted in non-decreasing order. Key Idea: Even though the input array is sorted, squaring can make negative numbers larger. So we use the two-pointer approach: One pointer at the start (left), one at the end (right). Compare absolute values, place the larger square at the end of the result array, and move pointers accordingly. #LeetCode #Java #DSA #TwoPointers #CodingJourney #ProblemSolving

🚀 Day 43 of #100DaysOfLeetCode Today's problem: LeetCode #160 – Intersection of Two Linked Lists 💡 Concept:  Find the node where two singly linked lists intersect.  Used the Two Pointer Approach — a smart and efficient way to solve this without using extra memory. 🧠 Logic:  Move both pointers through the lists.  When one pointer reaches the end, switch it to the other list’s head.  They’ll either meet at the intersection node or end up as null together. ✅ Complexity:  Time – O(n + m)  Space – O(1) 💬 Takeaway:  Sometimes, the best solutions come from balancing the path — literally!  Understanding how pointers sync up teaches a lot about memory references and linked list behavior. #LeetCode #CodingChallenge #Java #DataStructures #TwoPointerTechnique #ProblemSolving #LinkedLists

🚀 Day 43 of #100DaysOfLeetCode Today's problem: LeetCode #160 – Intersection of Two Linked Lists 💡 Concept:  Find the node where two singly linked lists intersect.  Used the Two Pointer Approach — a smart and efficient way to solve this without using extra memory. 🧠 Logic:  Move both pointers through the lists.  When one pointer reaches the end, switch it to the other list’s head.  They’ll either meet at the intersection node or end up as null together. ✅ Complexity:  Time – O(n + m)  Space – O(1) 💬 Takeaway:  Sometimes, the best solutions come from balancing the path — literally!  Understanding how pointers sync up teaches a lot about memory references and linked list behavior. #LeetCode #CodingChallenge #Java #DataStructures #TwoPointerTechnique #ProblemSolving #LinkedLists

#100DaysOfCode – Day 68 String Manipulation Problem 1:- Largest Odd Number in String Task:- Given a numeric string, return the largest-valued odd number (as a substring) or an empty string if none exists. Example: Input: num = "35427" → Output: "35427" My Approach: Started scanning the string from right to left. The first odd digit encountered marks the end of the required substring. Returned the substring from start to that index. Time Complexity:- O(N) Space Complexity:- O(1) Sometimes, it’s not about complex algorithms just a small logical observation can lead to an efficient solution. #takeUforward #100DaysOfCode #Java #ProblemSolving #LeetCode #CodeNewbie #StringManipulation #LogicBuilding #CleanCode

NeetCode 20 | LeetCode #567 | Permutation in String Applied sliding window with character frequency tracking to efficiently check for permutations in a string. A great exercise in combining logic, hash maps, and window optimization. #NeetCode #LeetCode #Java #DSA #Algorithms #SlidingWindow #ProblemSolving #CodingJourney

📌 Day 3/100 - Remove Duplicates from Sorted Array (LeetCode 26) 🔹 Problem: Given a sorted array, remove the duplicates in-place so that each element appears only once and return the new length. You must modify the array without using extra space for another array. 🔹 Approach: I used a simple counting-based approach: Iterate through the array using a single loop. If the current element is the same as the next, skip it. Otherwise, place it at the current count index and increment count. Finally, return count as the number of unique elements. 🔹 Key Learning: Practiced in-place array modification efficiently without extra space. Improved understanding of loop-based filtering logic. Realized that sometimes the simplest linear approach works best! Consistency compounds — each problem adds a new layer of confidence! 🚀#100DaysOfCode #LeetCode #Java #ProblemSolving #Array #DSA #TwoPointers