Day 95/200 – LeetCode Challenge. Problem: Remove Duplicates from Sorted List II (Java) Today’s focus was eliminating all duplicate values from a sorted linked list while keeping only distinct nodes. Implemented an efficient two-pointer approach with a dummy node to handle edge cases cleanly. Linked lists require careful pointer management. Dummy nodes simplify boundary conditions. One-pass solution ensures optimal performance. Continuing the 200-day journey, one problem at a time. #LeetCode #Java #CodingChallenge #200DaysOfCode #ProblemSolving
Remove Duplicates from Sorted List II Java LeetCode Challenge
More Relevant Posts
-
#Day95 of #100DaysOfCode Extended the previous number-to-words program to handle multiple digits. The program takes a number as input and prints each digit in its word form. Focused on improving logic by combining loops, string conversion, and conditional handling. #Java #MiniProject #100DaysOfCode
To view or add a comment, sign in
-
-
🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation
To view or add a comment, sign in
-
-
Day 74 – Delete Node in a Linked List Solved a unique linked list problem where only the node to be deleted is given, without access to the head. Key Learnings: Learned a technique to delete a node without accessing the previous node Used value replacement and pointer adjustment instead of traditional deletion Strengthened understanding of in-place linked list manipulation #DSA #Java #LinkedList #ProblemSolving #CodingPractice #DataStructures
To view or add a comment, sign in
-
-
Day 42 of consistency 💻🔥 Solved Add Two Numbers using linked lists — a classic problem that really tests understanding of pointers, carry handling, and edge cases. Key takeaways: Handling carry efficiently is crucial Dummy nodes make list problems much cleaner Iterative approach keeps space optimal Not the fastest runtime yet, but improvement is a process — optimizing step by step. Consistency > Perfection. #Day42 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode
To view or add a comment, sign in
-
-
Day 60 - Remove Duplicates from Sorted List Worked on removing duplicate elements from a sorted linked list. Approach: • Traverse the list once • Compare current node with next node • Skip duplicate nodes by adjusting pointers Time Complexity: O(n) Space Complexity: O(1) #Day60 #LeetCode #Java #LinkedList #CodingPractice #DSA #TechJourney
To view or add a comment, sign in
-
-
Day 69 - Remove Linked List Elements Working with linked lists to efficiently remove all nodes with a specific value using a dummy node approach. Time Complexity: O(n) Space Complexity: O(1) #Day69 #LeetCode #Java #LinkedList #DSA #ProblemSolving #CodingJourney
To view or add a comment, sign in
-
-
A small habit that saved me countless debugging hours: Always add meaningful logs. Not just: “Error occurred” But: “Payment service failed while calling order service – orderId:123” Future you (or another developer) will thank you. #Java #CleanCode #SoftwareDevelopment
To view or add a comment, sign in
-
🚀 Day 95/100 – LeetCode Challenge ✅ 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: 𝐒𝐞𝐭 𝐌𝐢𝐬𝐦𝐚𝐭𝐜𝐡 Today’s problem was about identifying a 𝐝𝐮𝐩𝐥𝐢𝐜𝐚𝐭𝐞 number and a 𝐦𝐢𝐬𝐬𝐢𝐧𝐠 number in a given array that should ideally contain values from 1 to n. 💡𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬: Understand frequency/counting approach Identify duplicate (count = 2) and missing (count = 0) Simple brute-force solution works but can be optimized Time Complexity: O(n²) 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Iterate from 1 → n Count occurrences of each number in the array If count = 2 → duplicate If count = 0 → missing #Day95 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving #KeepGoing
To view or add a comment, sign in
-
-
#Day85 of #100DaysOfCode Focused on practicing array-based problems with an emphasis on logic and efficiency. Worked on: * Moving zeros to the end of an array * Checking if an array is a palindrome * Calculating the sum of even elements #Java #Arrays #100DaysOfCode
To view or add a comment, sign in
-
Explore content categories
- Career
- Productivity
- Finance
- Soft Skills & Emotional Intelligence
- Project Management
- Education
- Technology
- Leadership
- Ecommerce
- User Experience
- Recruitment & HR
- Customer Experience
- Real Estate
- Marketing
- Sales
- Retail & Merchandising
- Science
- Supply Chain Management
- Future Of Work
- Consulting
- Writing
- Economics
- Artificial Intelligence
- Employee Experience
- Workplace Trends
- Fundraising
- Networking
- Corporate Social Responsibility
- Negotiation
- Communication
- Engineering
- Hospitality & Tourism
- Business Strategy
- Change Management
- Organizational Culture
- Design
- Innovation
- Event Planning
- Training & Development