Day 9 of 100DaysOfCode: Solving Leetcode Product of Array Except Self

Today’s problem: Merge Two Sorted Lists 🔗 Problem: Given two sorted linked lists, merge them into one sorted list and return it. Example: Input: list1 = [1,2,4], list2 = [1,3,4] Output: [1,1,2,3,4,4] Approach I used: ✅ Create a dummy node to simplify pointer operations. ✅ Use a pointer (tail) to build the merged list by comparing the heads of both lists. ✅ Append the smaller node each time and move forward. ✅ When one list ends, attach the remaining nodes of the other. A clean iterative solution that keeps the space usage minimal (O(1) extra space). ⚡

🚀 Day 57/100 of #100DaysOfLeetCode Today’s problem was 1323. Maximum 69 Number 💡 The task was to maximize a given positive integer (consisting only of digits 6 and 9) by changing at most one digit — specifically, changing a 6 to a 9 to get the largest possible number. 🔑 Key Takeaways: Learned how to efficiently manipulate numbers as strings. #LeetCode #Day57 #Java #100DaysOfCode #ProblemSolving #CodingChallenge #LearningEveryday

🚀 Day 7️⃣7️⃣ of #100DaysOfCode Solved LeetCode Problem #1437 — Check If All 1’s Are at Least K Places Away 🔍✨ 🔧 Concept: Given a binary array, ensure that every 1 is separated by at least K zeros. The solution simply tracks the previous index of 1 and checks the distance. 🧠 Key Idea: Scan the array once When a 1 is found, verify: 👉 currentIndex - previousIndex > k If any pair violates this, return false ⚡ Efficient: Time Complexity: O(n) Space Complexity: O(1) Another clean and efficient logical problem solved. #LeetCode #Java #Arrays #ProblemSolving #DSA #CodingJourney #100DaysOfCode #DeveloperLife #Motivation

💻 Day 76 of #100DaysOfCoding Today’s problem was “Check If Digits Are Equal in String After Operations I” (LeetCode – Easy). The task was to repeatedly sum each pair of consecutive digits (mod 10) until the string becomes exactly two digits long and then check if both are equal. I used a simple and clear approach with a while loop and StringBuilder, updating the string in each iteration until it reduced to two digits. It’s a great example of how small logical patterns can make an otherwise repetitive problem neat and efficient ✨ #100DaysOfCoding #LeetCode #CodingJourney #ProblemSolving #Java #CodeEveryday #DSA #WomenInTech #SoftwareEngineering #LearningInPublic

Day 2 of #251DaysOfDSA 🚀 Problem: Palindrome Number Time Complexity: O(log₁₀n) Space Complexity: O(1) ✨ What I learned: I learned how to check if a number reads the same backward as forward without converting it entirely into a string. The key idea is to reverse half of the number and compare it with the remaining half — simple yet efficient! #LeetCode #DSA #Java #251DaysOfDSA #CodingJourney #ProblemSolving #Consistency

#100DaysOfCode – Day 80 Reverse Linked List Problem: Given the head of a singly linked list, reverse the list and return the reversed version. Example: Input: [1, 2, 3, 4, 5] Output: [5, 4, 3, 2, 1] My Approach: Used an iterative method to reverse the list efficiently by manipulating pointers. Initialized three pointers prev, temp, and front. Iteratively reversed each node’s direction until the end of the list was reached. Returned prev as the new head of the reversed list. Time Complexity: O(N) Space Complexity: O(1) Reversing a linked list might look tricky at first, but once you understand pointer manipulation it’s pure logic and flow. #takeUforward #100DaysOfCode #DSA #Java #ProblemSolving #LeetCode #LinkedList #Pointers #CodeNewbie

💡 LeetCode #344 — Reverse String Today I solved LeetCode Problem 344: Reverse String 🔁 Problem Summary: Write a function that reverses a string. The input string is given as a character array s, and you must reverse it in-place (without using extra space). Key Idea: Use the two-pointer technique 👈👉 Initialize one pointer at the start (left) and another at the end (right). Swap the characters at both pointers and move them toward the center. #LeetCode #Java #DSA #TwoPointers #CodingJourney #ProblemSolving

🌳 Day 46 of #LeetCodeJourney Problem: 100. Same Tree ✅ Today’s challenge was about comparing two binary trees — checking if they’re structurally identical and have the same node values. A neat recursive problem that refreshes the basics of tree traversal and recursion! 💡 Key takeaway: If you can break a problem down into smaller identical subproblems — recursion will always have your back. #LeetCode #100DaysOfCode #Java #DSA #BinaryTree #Recursion #CodingJourney

✨ Day 26 🚀– #LeetCode Challenge ✨ 📘 Topic: Contains Duplicates https://lnkd.in/gVKGx5uP Today’s problem was all about checking whether an array has duplicate elements or not. 🔍 Concept Used: HashSet (or HashMap) for quick lookup Focus on time efficiency (O(n) complexity) 💡 Key Takeaway: Using a Set helps in identifying duplicates efficiently since it stores only unique values. 🧠 Logic Summary: Traverse through the array Add elements to a set If an element already exists → duplicate found ✅ #Day26 #LeetCode #30DaysOfCode #ProblemSolving #Java #CodingJourney #ContainsDuplicates #DataStructures #HashSet

#100DaysOfCode – Day 85 Delete the Middle Node of a Linked List Problem Statement: Given the head of a singly linked list, delete the middle node and return the modified list. My Approach: Used two-pointer technique (slow & fast) to identify the middle node in a single traversal. Maintained a prev pointer to skip the middle node without extra space. Carefully handled the edge case when the list has only one node. Complexity: Time: O(N) Space: O(1) Linked lists are all about pointers and precision a small mistake (like an extra semicolon ";") can make a huge difference. Understanding how slow and fast pointers move gives deep insight into efficient traversal patterns. #LeetCode #100DaysOfCode #Java #LinkedList #ProblemSolving #takeUforward #CodingJourney #DataStructures #CodeNewbie