HEMAMALINI M’s Post

 💻 Day 53 of #100DaysOfCode – LeetCode #206: Reverse Linked List Today’s challenge was one of the most classic problems in data structures — reversing a linked list. This problem is a great way to strengthen your understanding of pointers and linked list manipulation. 🧩 Problem:  Given the head of a singly linked list, reverse it and return the new head. ⚙ Approach:  I used an iterative approach with three pointers:  prev, curr, and next. Store the next node Reverse the current node’s pointer Move forward until the list is reversed 🧠 Key Learning:  This problem really helped me visualize how pointers work in linked structures — a great exercise for understanding memory links and iterative logic. ✅ Time Complexity: O(n) ✅ Space Complexity: O(1) Here’s my simple Java solution: class Solution {   public ListNode reverseList(ListNode head) {     ListNode prev = null, curr = head;     while (curr != null) {       ListNode next = curr.next;       curr.next = prev;       prev = curr;       curr = next;     }     return prev;   } } 🔥 Every day, one step closer to mastering data structures!  #Day53 #LeetCode #Java #100DaysOfCode #CodingJourney #LinkedList #DSA #Programming #ReverseLinkedList

💻 Day 53 of #100DaysOfCode – LeetCode #206: Reverse Linked List Today’s challenge was one of the most classic problems in data structures — reversing a linked list.  This problem is a great way to strengthen your understanding of pointers and linked list manipulation. 🧩 Problem:  Given the head of a singly linked list, reverse it and return the new head. ⚙ Approach:  I used an iterative approach with three pointers:  prev, curr, and next. Store the next node Reverse the current node’s pointer Move forward until the list is reversed 🧠 Key Learning:  This problem really helped me visualize how pointers work in linked structures — a great exercise for understanding memory links and iterative logic. ✅ Time Complexity: O(n)  ✅ Space Complexity: O(1) Here’s my simple Java solution: class Solution {   public ListNode reverseList(ListNode head) {     ListNode prev = null, curr = head;     while (curr != null) {       ListNode next = curr.next;       curr.next = prev;       prev = curr;       curr = next;     }     return prev;   } } 🔥 Every day, one step closer to mastering data structures!  #Day53 #LeetCode #Java #100DaysOfCode #CodingJourney #LinkedList #DSA #Programming #ReverseLinkedList

🎯 Day 82 of #100DaysOfCode – LeetCode 137: Single Number II Today’s challenge was about identifying the unique number in an array where every element appears three times except for one. 🧩 Problem Statement: Given an integer array nums, where every element appears exactly three times except for one, find that single element that appears only once. Example: Input: [2,2,3,2] Output: 3 💭 Approach 1: Using HashMap At first, I solved this using a HashMap to count the frequency of each number. Traverse the array and store each element’s occurrence. Return the element whose count equals 1. 🕒 Time Complexity: O(n) 💾 Space Complexity: O(n) 🚀 Key Takeaway: Sometimes, the best way to optimize is not by adding more data structures, but by understanding how data behaves at the bit level. #100DaysOfCode #Day82 #LeetCode #Java #ProblemSolving #DSA #CodingJourney #BitManipulation #KeepLearning

💼 Day 54 of #100DaysOfCode Challenge 🚀 📘 Problem: LeetCode #217 – Contains Duplicate  🧠 Topic: HashSet | Array | Data Structures 🔍 Problem Statement: Given an integer array nums, return true if any value appears at least twice in the array, and false if every element is distinct. ⚙ Approach: Used a HashSet to check for duplicates efficiently. Traverse through each number in the array. If the number already exists in the set → return true. Otherwise, add it to the set and continue. If loop ends with no duplicates → return false. 💻 Java Solution: import java.util.HashSet; public class Solution {   public boolean containsDuplicate(int[] nums) {     HashSet<Integer> set = new HashSet<>();     for (int num : nums) {       if (set.contains(num)) {         return true;        }       set.add(num);     }     return false;    } } ⏱ Complexity Analysis: Time: O(n) Space: O(n) 💬 Takeaway: Simple but powerful — using the right data structure like HashSet can make problems easy and efficient to solve. ⚡ #100DaysOfCode #Day54 #LeetCode #Java #CodingChallenge #DSA #HashSet #ProblemSolving #Programmer #TechJourney

💼 Day 54 of #100DaysOfCode Challenge 🚀 📘 Problem: LeetCode #217 – Contains Duplicate  🧠 Topic: HashSet | Array | Data Structures 🔍 Problem Statement: Given an integer array nums, return true if any value appears at least twice in the array, and false if every element is distinct. ⚙ Approach: Used a HashSet to check for duplicates efficiently. Traverse through each number in the array. If the number already exists in the set → return true. Otherwise, add it to the set and continue. If loop ends with no duplicates → return false. 💻 Java Solution: import java.util.HashSet; public class Solution {   public boolean containsDuplicate(int[] nums) {     HashSet<Integer> set = new HashSet<>();     for (int num : nums) {       if (set.contains(num)) {         return true;        }       set.add(num);     }     return false;    } } ⏱ Complexity Analysis: Time: O(n) Space: O(n) 💬 Takeaway: Simple but powerful — using the right data structure like HashSet can make problems easy and efficient to solve. ⚡ #100DaysOfCode #Day54 #LeetCode #Java #CodingChallenge #DSA #HashSet #ProblemSolving #Programmer #TechJourney

💼 Day 54 of #100DaysOfCode Challenge 🚀 📘 Problem: LeetCode #217 – Contains Duplicate 🧠 Topic: HashSet | Array | Data Structures 🔍 Problem Statement: Given an integer array nums, return true if any value appears at least twice in the array, and false if every element is distinct. ⚙️ Approach: Used a HashSet to check for duplicates efficiently. Traverse through each number in the array. If the number already exists in the set → return true. Otherwise, add it to the set and continue. If loop ends with no duplicates → return false. 💻 Java Solution: import java.util.HashSet; public class Solution { public boolean containsDuplicate(int[] nums) { HashSet<Integer> set = new HashSet<>(); for (int num : nums) { if (set.contains(num)) { return true; } set.add(num); } return false; } } ⏱ Complexity Analysis: Time: O(n) Space: O(n) 💬 Takeaway: Simple but powerful — using the right data structure like HashSet can make problems easy and efficient to solve. ⚡ #100DaysOfCode #Day54 #LeetCode #Java #CodingChallenge #DSA #HashSet #ProblemSolving #Programmer #TechJourney

🚀 Just implemented a custom Vector<T> class in Java — a dynamic array structure that grows automatically as elements are added. It supports key operations like push_back, pop_back, front, back, clear, and more — all while practicing the concepts of generics, dynamic memory allocation, and data encapsulation. Rebuilding core data structures from scratch is a great way to understand how tools like ArrayList work internally. 💻 #Java #DSA #LearningByDoing #CodingJourney #DataStructures Nohit Singh,Shreyash Khedekar,Anshuman Singh,Arpit Jain

Sharing the simple script I developed for my PhD research: an interactive Java program that generates problem instance files and matching solution files for vehicle routing / dropoff coordination experiments. It currently: Supports instance templates: 2, 3, 5, 9, 12, 15, 18, 21 Prompts for integer scenario parameters (vehicles, customers, depots, deployment time, capacity, demand/frequency, start/end depot) Produces two files per run: FINALDATA/p{N}.txt and FINALSOLUTION/pr{N}.txt (paths currently hard-coded) Includes ready-made solution routes and objective values per template (useful for benchmarking or as fixtures) #VehicleRouting #VRP #Logistics #RoutingAlgorithms https://lnkd.in/giFPQiNC

Day 41 of 160 Days of DSA 🚀 Topic: Set Matrix Zeroes Today, I learned a valuable lesson while solving the Set Matrix Zeroes problem — sometimes the order in which we update data matters more than the logic itself. The core idea of the problem: If any element in the matrix is 0, we have to set its entire row and column to 0. At first, I tried to directly modify the matrix during traversal. But this caused new zeroes to appear, which led to overwriting more rows/columns than required. So the correct approach is: First pass → Scan the matrix and record which rows and columns should be zeroed. Second pass → Set those rows and columns to zero. This prevents us from modifying the matrix while still evaluating it. Key Takeaway: When transforming a data structure based on its own values, sometimes the right answer requires separating detection from modification. #100DaysOfCode #geeksforgeeks #dsa #java #arrays #learningEveryday #codingjourney #logicbuilding