Determine Matrix Rotation in Java

Day: 94/365 📌 LeetCode POTD: Minimum Distance Between Three Equal Elements I Easy Key takeaways/Learnings from this problem: 1. This problem is a nice reminder that tracking indices smartly is often more important than the values themselves. 2. Keeping the last few occurrences of each element helps you quickly check valid triplets without brute force. 3. It shows how a simple hashmap or array tracking can turn an O(n³) idea into something efficient. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 95/365 📌 LeetCode POTD: Minimum Distance Between Three Equal Elements II Medium Key takeaways/Learnings from this problem: 1. This one highlights how tracking indices smartly is more important than checking all triplets brute-force. 2. Using previous occurrences helps you narrow down valid triples quickly instead of re-scanning the array. 3. Big takeaway: problems with “equal elements” often reduce to index management + pattern observation. 4. Also a good reminder that optimizing from O(n³) thinking to near O(n) comes down to storing the right info while iterating. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 78/365 📌 LeetCode POTD: Equal Sum Grid Partition I Medium Key takeaways/Learnings from this problem: 1. This problem highlights how prefix sums make grid partitioning super efficient instead of recalculating sums again and again. 2. The key is trying splits smartly (row-wise or column-wise) and checking if both parts can balance out. 3. It’s a good reminder that many 2D problems become easier when you reduce them to 1D cumulative sums. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day 42 of consistency 💻🔥 Solved Add Two Numbers using linked lists — a classic problem that really tests understanding of pointers, carry handling, and edge cases. Key takeaways: Handling carry efficiently is crucial Dummy nodes make list problems much cleaner Iterative approach keeps space optimal Not the fastest runtime yet, but improvement is a process — optimizing step by step. Consistency > Perfection. #Day42 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

Day: 76/365 📌 LeetCode POTD: Maximum Non Negative Product in a Matrix Medium Key takeaways/Learnings from this problem: 1. This problem shows why tracking only max isn’t enough—you need both max and min products because negatives can flip the result. 2. Classic DP twist where each cell depends on top and left, but with sign handling making it interesting. 3. Big takeaway: whenever multiplication + negatives are involved, always think about dual states (min & max). #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

🚀 Day #82/100 – 𝐑𝐞𝐦𝐨𝐯𝐞 𝐄𝐥𝐞𝐦𝐞𝐧𝐭 (LeetCode) Today’s problem was Remove Element — a simple yet important question to understand in-place array manipulation. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We don’t need extra space! We can solve this using a two-pointer approach efficiently.  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We overwrite unwanted elements and keep only the valid ones at the beginning of the array. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize k = 0 Traverse array: If element ≠ val → place it at index k Increment k Return k (new length) ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(n) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(1) #Day82 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Solved a problem where we need to check if two strings can be made equal using a special operation. The rule is: you can swap characters only if the distance between their positions is even. So basically, characters at even indices can only swap among themselves, and same for odd indices. Idea: Instead of actually swapping, I just counted characters separately for even and odd positions in both strings. If both match, then it’s possible — otherwise not. Simple concept, but interesting twist! 😊 #LeetCode #DSA #Java #ProblemSolving #CodingJourney

Solved LeetCode 2839 – Check if Strings Can be Made Equal With Operations I today. In simple terms, we are given two strings and allowed to swap characters only at positions that are 2 indices apart (like index 0 with 2, 1 with 3). So instead of trying all swaps, the idea is to separate characters at even and odd positions and check if both strings have the same characters in those positions. If the frequency of characters matches for even and odd indices separately, then the strings can be made equal. A simple yet interesting problem that improves thinking around patterns and constraints. #LeetCode #DSA #Java #ProblemSolving #CodingJourney

🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation