Linked List Cycle Detection with Two Pointers

Day 14/100 – LeetCode Challenge Problem: Linked List Cycle II Today’s problem focused on detecting the node where a cycle begins in a linked list. Approach: Used Floyd’s Cycle Detection Algorithm (Slow & Fast pointers). Move slow by one step and fast by two steps. If they meet, a cycle exists. Reset slow to head. Move both slow and fast one step at a time. The node where they meet again is the starting node of the cycle. Complexity: Time: O(n) Space: O(1) Concepts Practiced: Linked List traversal Floyd’s cycle detection Pointer manipulation #100DaysOfCode #LeetCode #DSA #Java #LinkedList #ProblemSolving #CodingJourney

Day 13/100 – LeetCode Challenge Problem: Linked List Cycle Today’s problem was about detecting whether a cycle exists in a linked list. Approach: Used Floyd’s Cycle Detection Algorithm (Tortoise and Hare). slow pointer moves one step fast pointer moves two steps If a cycle exists, both pointers will eventually meet If fast reaches null, the list has no cycle Complexity: Time: O(n) Space: O(1) Concepts Practiced: Linked List traversal Two-pointer technique Cycle detection algorithm #100DaysOfCode #LeetCode #DSA #Java #LinkedList #ProblemSolving #CodingJourney

Day 7/100 – LeetCode Challenge Problem: Palindrome Number Today’s problem focused on number manipulation without converting the integer into a string. Approach: If number is negative → return false Reverse the digits using modulo (% 10) and division (/ 10) Compare reversed number with original number Logic Used: Extract last digit: rem = x % 10 Build reversed number: rev = rev * 10 + rem Remove last digit: x = x / 10 Complexity: Time: O(log₁₀ n) Space: O(1) Key takeaway: Problems involving digits can often be solved mathematically without string conversion. #100DaysOfCode #LeetCode #DSA #Java #ProblemSolving #CodingJourney

🚀 Day 42 of #100DaysOfCode 🌱 Topic: Linked List / Two Pointers ✅ Problem Solved: LeetCode 82 – Remove Duplicates from Sorted List II 🛠 Approach: Used a dummy node to simplify handling edge cases where the head might be removed. When two consecutive nodes had the same value, stored that value. Skipped all nodes with that duplicate value using a loop. Linked the previous node to the next distinct node. Continued traversal until reaching the end. This ensures that only unique elements remain in the sorted list. #100DaysOfCode #Day42 #DSA #LinkedList #TwoPointers #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

Day 42 of DSA 🚀 | Alternating Binary String Today’s problem: Minimum Changes To Make Alternating Binary String 🔹 Task: Convert a binary string into an alternating string (0101... or 1010...) using the minimum number of flips. 💡 Key Insight: Only two valid patterns exist: 010101... 101010... Count mismatches with one pattern and compute the other using: min(count, n - count) 📌 Example s = "0100" → Output = 1 ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) Small problem, but a great reminder that recognizing patterns can simplify the solution drastically. #DSA #Java #LeetCode #ProblemSolving #LearningInPublic #Consistency

🧠 Day 30 / 100 – DSA Practice Solved Remove Duplicates from Sorted List on LeetCode 🔗✅ 🔹 Problem Insight: Given a sorted linked list, remove duplicates so each element appears only once. 🔹 Approach: Used a single pointer traversal: Compared current node with next node Skipped duplicate nodes by updating links Leveraged the fact that the list is already sorted 🔹 Complexity: Time → O(n) Space → O(1) 💯 Result: ✔️ All test cases passed ⚡ Runtime: 0 ms (Beats 100%) Consistency builds confidence 💪🚀 #Day30 #100DaysOfCode #LeetCode #Java #DSA #LinkedList #CodingJourney #ProblemSolving

#day328 of #1001daysofcode problem statement (1758): Minimum Changes To Make Alternating Binary String An alternating binary string can only follow two patterns: "010101..." or "101010...". I counted the number of changes required to convert the string into both patterns and returned the minimum of the two. ⏱Time Complexity: O(n) 🧠Space Complexity: O(1) Consistency check ✅ One LeetCode problem a day to sharpen problem-solving skills. #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

Day 52 of #100DaysOfLeetCode 🚀 Solved Balanced Binary Tree (Easy) 🌳 Learned how to efficiently check if a binary tree is height-balanced using a single DFS traversal. Instead of recalculating heights multiple times (O(n²)), optimized it to O(n) by combining height calculation with balance checking. 💡 Key takeaway: Return -1 early when imbalance is detected to avoid unnecessary computations. #LeetCode #DSA #Java #CodingJourney

#day333 of #1001daysofcode problem statement (0257): Binary Tree Paths 💡 Approach: Used DFS recursion to explore all root-to-leaf paths in the binary tree. While traversing, I kept building the path string. When a leaf node is reached, the complete path is added to the result list. Example path format: 1->2->5 ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(h) — recursion stack (h = height of the tree) Consistency in solving one problem every day 📈 #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

🚀Day 11 of #Leetcode75 LeetCode #392 – Is Subsequence | Two Pointer Approach Today I solved “Is Subsequence” (Easy) and revised an important concept — the Two Pointer Technique. 🔎 Problem Summary: Given two strings s and t, determine whether s is a subsequence of t. A subsequence maintains the relative order of characters but doesn’t need to be continuous. 💡 Key Insight: Instead of using extra space or complex logic, we can solve this efficiently using two pointers: ✔ One pointer for s ✔ One pointer for t ✔ Move forward and match characters in order ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) This problem reinforces how powerful simple logic can be when applied correctly. Consistency > Complexity 💪 Excited to keep improving step by step! #LeetCode #DSA #Java #CodingPractice #ProblemSolving #100DaysOfCode #TechJourney