Day 26: Two Pointers and Prefix Sum Practice

#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 47 of #100DaysOfCode 🌱 Topic: Arrays / HashMap ✅ Problem Solved: LeetCode 260 – Single Number III 🛠 Approach: Used a HashMap to track frequency of elements. Traversed the array and stored counts of each number. Iterated through the map to find elements with frequency 1. Stored those elements in the result array. This approach is straightforward but uses extra space. #100DaysOfCode #Day47 #DSA #Arrays #HashMap #BitManipulation #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

🚀 Day 55 of #100DaysOfCode Solved 147. Insertion Sort List on LeetCode 🔗 🧠 Key Insight: We apply the classic Insertion Sort, but on a linked list instead of an array. The challenge is handling pointer manipulation efficiently. ⚙️ Approach: 1️⃣ Create a dummy node to act as the start of the sorted list 2️⃣ Traverse the original list node by node 3️⃣ For each node: Find its correct position in the sorted part Insert it there by updating pointers 🔁 This builds a sorted list incrementally ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Sorting #Java #InterviewPrep #CodingJourney

Day: 57/365 📌 LeetCode POTD: Special Positions in a Binary Matrix Easy Key takeaways/Learnings from this problem: 1. This one shows how precomputing row and column counts makes checking each cell super easy. 2. Instead of re-scanning the whole row and column every time, count once and reuse smartly. 3. Good reminder that brute force can often be optimized with a tiny bit of preprocessing. 4. Clean implementation and careful condition checking matter more than complex algorithms here. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

#Day39 of #365DaysOfCode Solved Set Matrix Zeroes on LeetCode. Implemented an in-place matrix approach using the first row and column as markers to track zero positions. ✨ Lesson: Smart space optimization and careful traversal help solve matrix problems efficiently. #CodingJourney #DSA #Java #LeetCode #Consistency

🚀 Day 39 of #100DaysOfCode Solved 80. Remove Duplicates from Sorted Array II on LeetCode 🔢 🧠 Key Insight: The array is already sorted, and we need to ensure that each element appears at most twice, modifying the array in-place. ⚙️ Approach: 🔹Maintain a pointer i representing the position to place the next valid element 🔹Start iterating from index 2 🔹For each element nums[j], compare it with nums[i] 🔹If they are different, place the element at nums[i + 2] and move the pointer forward This ensures that no element appears more than twice while maintaining the sorted order. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #Arrays #TwoPointers #Java #ProblemSolving #InterviewPrep #LearningInPublic

Day 16/100 – LeetCode Challenge Problem: Merge Sorted Array Today’s problem involved merging two sorted arrays into one sorted array. Approach: Created a temporary array of size m + n Used two pointers to compare elements from both arrays Inserted the smaller element into the new array Copied remaining elements if any array still had values Finally copied the merged result back into nums1 Complexity: Time: O(m + n) Space: O(m + n) Concepts Practiced: Two-pointer technique Array traversal Merging sorted arrays #100DaysOfCode #LeetCode #DSA #Java #Arrays #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 35 – Plus One Worked on a problem where a large integer is represented as an array of digits, and the task was to increment the number by one. Key Learnings: Traversing arrays from right to left to simulate digit addition Handling carry when a digit becomes 10 Managing edge cases where all digits are 9 by creating a new array #DSA #Java #Arrays #ProblemSolving #CodingPractice #LeetCode

#day332 of #1001daysofcode problem statement (0226): Invert Binary Tree The idea is to recursively swap the left and right child of every node in the tree. ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(h) — recursion stack (h = height of tree) #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode