Linked List Components LeetCode Challenge

leetcode thoughts Learn to make proper recall while revisiting your solution Creating comment # when you got compilation error while running your code # when you do optimization # when you got Submission error (edge case error, time and space complexity issues) Create solid understanding while revisiting the code #ThoughtsFromExperience #leetcode #pattern_understanding #EffectiveLearning

🚀 Day 80 of #100DaysOfCode Today, I solved LeetCode 61 – Rotate List, a problem focused on linked list manipulation and efficient pointer handling. 💡 Problem Overview: Given the head of a linked list, the task is to rotate the list to the right by k places. 🧠 Approach: ✔️ Calculated the length of the linked list ✔️ Connected the tail to the head to form a circular list ✔️ Found the new tail position using k % length ✔️ Broke the cycle to get the rotated list This approach avoids unnecessary rotations and ensures optimal performance. ⚡ Key Takeaways: Linked list problems require strong pointer manipulation Converting problems into circular structures can simplify logic Modulo operation helps handle large values of k efficiently 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 🎯 Day 80 Insight: Consistency over time builds confidence in solving even complex problems efficiently. On to Day 100 🚀 #LeetCode #100DaysOfCode #DSA #LinkedList #ProblemSolving #CodingJourney #SoftwareDevelopment #Consistency #InterviewPrep

🚀 Day 17/100 — LeetCode Challenge Solved Reverse Linked List At first, it feels tricky because we need to change the direction of links without losing the list. 💡 Approach (Iterative): 1) Use three pointers: prev, curr, next 2) Store the next node 3) Reverse the link (curr → prev) 4) Move all pointers forward 👉 Repeat until the list is fully reversed 💡 Key Insight: We’re not just moving through the list, we’re rebuilding it in reverse order. 🧠 Time Complexity: O(n) 💾 Space Complexity: O(1) 💡 What I learned: Linked List problems are less about logic, and more about careful pointer handling. One small mistake = broken list. #LeetCode #DSA #100DaysOfCode #Cpp #LinkedList #CodingJourney

Day 50 of my #50DaysOfCode challenge is done ✅🎉 📌 Problem Solved Reorder Linked List We were given a linked list. Task was to reorder it like: L0 → Ln → L1 → Ln-1 → L2 → Ln-2 … Not just reverse. Rearrange in a specific pattern. 💻 Approach 🔹️Find the middle of the linked list. 🔹️Reverse the second half. 🔹️Merge both halves alternately. Step by step: 🔹️Use slow and fast pointers to find middle 🔹️Reverse second half of list 🔹️Merge nodes one by one Careful pointer handling needed. 📊 Complexity Analysis Time Complexity: O(n) Space Complexity: O(1) 📚 What I learned today: ▫️Linked list problems often combine multiple steps. ▫️Finding middle + reversing + merging is a common pattern. ▫️Pointer manipulation needs careful attention. ▫️Breaking problem into parts makes it manageable. 🎯 50 Days Completed Started with basic patterns. Reached linked lists, stacks, recursion, sliding window. Some days were easy. Some were confusing. But I showed up every day. That’s the biggest win. This challenge was not just about coding. It was about consistency and discipline. #50DaysOfCode #CodingChallenge #Consistency #LearningInPublic

Day 18 / 100 Days of Code Challenge 💻🔥 Solved LeetCode 155 — Min Stack 📊 🔍 Problem • Design a stack that supports standard operations: push, pop, top • Additionally, implement getMin() to return the minimum element • All operations must run in O(1) time ⚙️ Approach (Two Stack Technique) • Use one stack to store all elements • Use another stack to keep track of minimum values • While pushing, store the current minimum • While popping, update the minimum accordingly 💡 Key Learning • Maintaining extra information in data structures • Efficient tracking of minimum without traversal • Understanding how to achieve constant time operations ⏱ Complexity • Time: O(1) for all operations ⚡ • Space: O(n) 📦 Consistency continues 🚀 #100DaysOfCode #LeetCode #DSA #Stack #ProblemSolving #Consistency

🚀 Day 64/100 – LeetCode Challenge. 🔍 Problem: Valid Parentheses. Today’s problem was all about checking whether a string of brackets is valid or not. Sounds simple… but the trick lies in handling it efficiently! 💡 Approach Used: Stack (LIFO) 👉 Idea: Push opening brackets → ( { [ On closing bracket → check top of stack If match → pop Else → invalid ❌ 🧠 Key Insight: Stack helps maintain the correct order of brackets. The last opened bracket must be the first one to close. ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(n) 📌 Takeaway: Whenever you see problems involving matching pairs / nested structures, think of using a stack! #100DaysOfCode #DSA #LeetCode #CodingJourney

🚀 Day 76 of #100DaysOfCode Today, I solved LeetCode 18 – 4Sum, a classic problem that extends the two-pointer technique to higher complexity. 💡 Problem Overview: Given an array, the task is to find all unique quadruplets that sum up to a target value. 🧠 Approach: ✔️ Sorted the array to simplify processing ✔️ Fixed two elements using nested loops ✔️ Applied the two-pointer technique for the remaining two elements ✔️ Carefully handled duplicates to ensure unique quadruplets This approach efficiently reduces unnecessary computations compared to brute force. ⚡ Key Takeaways: Sorting + Two Pointers is a powerful combination Avoiding duplicates is crucial in combination problems Breaking a complex problem into smaller parts simplifies logic 📊 Complexity Analysis: Time Complexity: O(n³) Space Complexity: O(1) (excluding output) Consistently pushing boundaries and solving more complex problems 🚀 #LeetCode #100DaysOfCode #DSA #TwoPointers #ProblemSolving #CodingJourney #SoftwareDevelopment #InterviewPrep

🚀 Day 14/100 — LeetCode Challenge Solved Delete Node in a Linked List The twist? 👉 You are not given the head of the list At first, I thought deletion wouldn’t be possible without the previous node. But the trick is: 1) Copy the value of the next node 2) Point current node to next->next 3) Effectively “delete” the next node instead 💡 Key Insight: Instead of deleting the given node directly, we shift values and bypass the next node. 🧠 Time Complexity: O(1) 💾 Space Complexity: O(1) 💡 What I learned: Sometimes problems are not about applying a pattern, but about thinking differently with given constraints. Switching from patterns to concepts now. #LeetCode #DSA #100DaysOfCode #Cpp #LinkedList #CodingJourney

Day 49 of my #50DaysOfCode challenge is done ✅ 📌 Problem Solved Remove All Adjacent Duplicates in String We were given a string s. Task was to remove all adjacent duplicates. If duplicates are removed, new duplicates may form. So we repeat until no duplicates remain. Example: "abbaca" → "ca" 💻 Approach (Using Stack) 🔹️Create an empty stack. 🔹️Traverse each character. 🔹️If stack is not empty and top == current character → pop 🔹️Else → push current character 🔹️At the end, build string from stack Duplicates get removed automatically. 📊 Complexity Analysis Time Complexity: O(n) Space Complexity: O(n) 📚 What I learned today: ▫️Stack helps in handling adjacent comparisons. ▫️Removing elements can create new patterns. ▫️Single pass solution is possible with stack. ▫️String problems often map to stack operations. Day 49 completed. Almost at 50 🚀 #50DaysOfCode #CodingChallenge #Consistency #LearningInPublic