LeetCode POTD: Flip Square Submatrix Vertically in Java

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 56 - Search a 2D Matrix Approached the problem by leveraging the sorted property of the matrix. Applied binary search over virtual 1D space instead of traversing row-wise. Time Complexity: O(log(m*n)) #Day56 #LeetCode #Java #BinarySearch #Matrix #DSA #CodingPractice #ProblemSolving

Day 64 - Intersection of Two Linked Lists Solved by using a two-pointer technique to find the intersection node of two linked lists. Instead of calculating lengths, both pointers traverse both lists to align automatically. Time Complexity: O(n + m) Space Complexity: O(1) #Day64 #LeetCode #Java #LinkedList #DSA #ProblemSolving #CodingJourney

✨ Day 38 of 90 – Pattern Mastery Journey 🧠 Pattern:Binary Triangle Pattern 💡 Approach: ✔ Used nested loops to control rows and columns ✔ Applied a simple condition `(i + j) % 2` to alternate values ✔ Printed ‘1’ when the sum is even, otherwise ‘0’ ✔ No extra variables needed — clean and efficient logic 🚀 This problem helped me understand how **mathematical conditions can simplify pattern logic**, making the code more optimized and readable. #PatternMasteryJourney #Java #CodingJourney #ProblemSolving

Day 72 of #100DaysOfCode Problem: Convert Sorted Array to Height-Balanced BST Today I learned how to efficiently convert a sorted array into a balanced Binary Search Tree using Divide & Conquer. Key Insight: Pick the middle element as the root to maintain balance. Recursively build: Left subtree from left half Right subtree from right half This ensures: Optimal height Faster search operations ⏱ Time Complexity: O(n) 📦 Space Complexity: O(log n) Consistency is the real game changer #DSA #Java #BinaryTree #LeetCode #CodingJourney

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #CodingJourney

Day 52 of #100DaysOfLeetCode 🚀 Solved Balanced Binary Tree (Easy) 🌳 Learned how to efficiently check if a binary tree is height-balanced using a single DFS traversal. Instead of recalculating heights multiple times (O(n²)), optimized it to O(n) by combining height calculation with balance checking. 💡 Key takeaway: Return -1 early when imbalance is detected to avoid unnecessary computations. #LeetCode #DSA #Java #CodingJourney

✨ Day 39 of 90 – Pattern Mastery Journey 🧠 Pattern: Concentric Square Number Pattern 💡 Approach: ✔ Created a square matrix of size (2*n - 1) ✔ Calculated distance from all four sides (top, bottom, left, right) ✔ Used the minimum distance to determine the value ✔ Applied formula → value = n - min(distance) ✔ Generated layer-by-layer decreasing pattern 🚀 This problem helped me understand **matrix-based thinking and how patterns can be formed using distance logic**. #PatternMasteryJourney #Java #CodingJourney #ProblemSolving

𝐃𝐚𝐲 87/100 – 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 Problem: 228. 𝐒𝐮𝐦𝐦𝐚𝐫𝐲 𝐑𝐚𝐧𝐠𝐞𝐬  Today I solved a problem where we need to summarize consecutive numbers in a sorted unique array into ranges. 🔑 𝐈𝐝𝐞𝐚: Traverse the array and keep extending the range while consecutive numbers continue. Once the sequence breaks, close the range and store it. 💡 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Start with the first element as start Move forward while nums[i] + 1 == nums[i+1] If range exists → "start->end" Else → single number "start" 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: Efficient single pass solution (O(n)) by grouping consecutive elements on the fly. #LeetCode #Java #ProblemSolving #DSA #100DaysOfCode #CodingJourney