30 Days of Code: Hashing Progress

#Day72 Of Problem Solving Solved today’s LeetCode Daily Challenge ✅. Problem: Check if Binary String Has at Most One Segment of Ones Sometimes, the simplest logic gives the best results. Instead of overthinking, I went with a clean approach—just checking if "01" exists in the string. If it does, that means more than one segment of 1’s… and that’s it! ✔️ All test cases passed ⚡ Runtime: 0 ms 📊 Beat 100% in performance This problem was a good reminder that not every solution needs complexity—clarity matters more. Consistency > Complexity 🚀 On to the next challenge! #LeetCode #DailyChallenge #Java #ProblemSolving #CodingJourney #100DaysOfCode #LinkedIn

🚀 Day 53 of #100DaysOfLeetCode Solved: Next Greater Element (Leetcode 496) Today’s problem was a great reminder of how powerful Monotonic Stack can be for optimization. 🔹 Approach: Used a stack to maintain decreasing elements Mapped each element to its next greater using a HashMap Reduced time complexity from O(n²) → O(n) 🔹 Key Learning: Small mistakes in conditions (like missing a !) can completely break logic. Attention to detail matters just as much as understanding the concept. 🔹 Complexity: Time: O(n + m) Space: O(n) Consistency > Perfection. Showing up daily and improving step by step. #LeetCode #DSA #Java #CodingJourney #ProblemSolving

Everything looked fine… Until the output surprised me 👀 Same key. Different values. But only one survived 🤯 That’s when I realized — HashMap doesn’t store duplicates the way we think 🧠 One small detail… And your data changes completely ⚠️ Many developers miss this in real projects. 🚨 Stop just watching tutorials… Real growth = Practice + Consistency 💯 🔥 Java Daily Practice ☕️ 👉 Join & start today 🔗 https://lnkd.in/gfhqgjGd 🚀 Here’s a quick challenge 👇 💬 What do you think the output will be? #Java #JavaDeveloper #HashMap #Collections #Programming #Debugging #BackendDeveloper #Coding #TechLearning #DeveloperTips #LinkedInIndia

🚀 Day 63 of #LeetCode Practice Solved: Count Equal and Divisible Pairs in an Array (Easy) 🔹 Problem Summary: Given an array and an integer k, find the number of pairs (i, j) such that: • nums[i] == nums[j] • (i * j) is divisible by k 🔹 Approach: • Used two nested loops to check all pairs • Compared elements to find equal values • Checked if (i * j) % k == 0 • Counted valid pairs 🔹 Key Learning: Brute-force helps build a strong foundation and improves problem understanding before optimization. 💡 Consistency, patience, and small daily improvements lead to big results over time. #LeetCode #DSA #Java #CodingJourney #Consistency #KeepLearning

After a long time, I solved a hard problem on LeetCode today. Small wins like this keep the learning journey exciting. 🚀 🚀 #Day82 of #100DaysOfCodeChallenge Today’s problem was a challenging one focused on Linked Lists and pointer manipulation. 📌 Problem: Reverse Nodes in k-Group (Hard) Given a linked list, the task is to reverse nodes in groups of size k. If the remaining nodes are fewer than k, they stay as they are. 🧠 Key Idea: Traverse the list and count k nodes. If k nodes exist, recursively process the remaining list. Reverse the current group of k nodes. Connect the reversed group with the result of the recursive call. Problems like this really test your understanding of how nodes connect and move inside a linked list. Consistency continues. One step closer to becoming better at problem solving every day. 💪 #100DaysOfCode #Day82 #LeetCode #DSA #LinkedList #Java #ProblemSolving #CodingJourney

From confusion to clarity solved a LeetCode Hard today. 🚀 🚀 #Day83 of #100DaysOfCodeChallenge Today’s challenge focused on Linked Lists and merging multiple sorted structures efficiently. 📌 Problem: Merge k Sorted Lists (Hard) You are given k sorted linked lists, and the goal is to merge them into one sorted linked list. 🧠 Approach: Start with an empty result list. Iterate through each linked list in the array. Merge the current result list with the next list using a helper merge function. The helper function merges two sorted linked lists using the two-pointer technique. This problem is a great way to strengthen understanding of handling multiple linked lists and maintaining sorted order. Consistency continues learning and improving every day. 💪 #100DaysOfCode #Day83 #LeetCode #DSA #LinkedList #Java #ProblemSolving #CodingJourney

🚀 #60DaysOfLeetCode – Day 32 Today’s problem was about removing all adjacent duplicates in a string. 🔹 Approach Used: Stack To efficiently handle consecutive duplicates, I used a stack-based approach: Traverse the string character by character. If the stack is not empty and the top element matches the current character, pop it (this removes the duplicate pair). Otherwise, push the current character onto the stack. Finally, build the result string from the stack and reverse it to maintain the correct order. 🔹 Key Learning: Using a stack helps simulate the process of removing adjacent duplicates in a single pass, achieving O(n) time complexity and avoiding unnecessary reprocessing. This problem highlights how stacks are powerful for handling sequential patterns and cancellations, especially in string manipulation problems. On to Day 33! 💻🔥 #LeetCode #DSA #Java #CodingChallenge #ProblemSolving #LearningInPublic #60DaysOfCode

Day 52 : Practicing Binary Trees on LeetCode 💡 Today’s live practical session in Alpha Plus 7.0 we solve some Binary Trees bases questions on LeetCode to test our logic. What I practiced today: ✅ Binary Tree Execution: Reinforced our core concepts by solving practical problems and tracing the exact flow of the nodes. ✅ Preorder Traversal: tackled the "Binary Tree Preorder Traversal" problem on LeetCode. ✅ Build the logic breakdown #BinaryTree #LeetCode #ProblemSolving #DSA #Java #SoftwareEngineering #100DaysOfCode #ApnaCollege

Day 33 of #geekstreak60 Today’s problem was about finding the longest cycle in a directed graph The key observation was that each node has at most one outgoing edge, so the graph behaves like chains and cycles. While traversing, I stored the step at which each node was visited. If I reach the same node again in the same path → cycle found. Cycle length = difference in steps. Interesting problem with a neat trick #geekstreak60 #npci #Day33 #GeeksforGeeks #DSA #Java #ProblemSolving #CodingJourney #Consistency

🌱 Day 31 of #100DaysOfCode 🚀 📍 Day 31 Progress Practiced problem-solving with focus on brute force vs optimized approaches ✅Solved problems: 📌Two Sum (unsorted array) Brute force approach using array iteration Optimized approach using HashMap 📌Container With Most Water Brute force approach using nested loops Optimized approach using the Two-Pointer technique #100DaysOfCode #DSA #TwoPointers #Hashing #Java #LearningInPublic #Consistency