Minimum Operations to Equalize Binary String with Java

Day: 71/365 📌 LeetCode POTD: Count Submatrices with Top-Left Element and Sum Less Than k Medium Key takeaways/Learnings from this problem: 1. Prefix sum is the hero here—once you precompute it, submatrix sum becomes instant lookup instead of recomputing every time. 2. Fixing the top-left corner simplifies things a lot, turning a 2D problem into something more manageable. 3. Instead of brute forcing all submatrices, you just expand and check smartly using precomputed sums. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

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: 72/365 📌 LeetCode POTD: Count Submatrices With Equal Frequency of X and Y Medium Key takeaways/Learnings from this problem: 1. The key idea is converting the problem into numbers (like +1 and -1) so equal frequency turns into a sum = 0 problem. 2. 2D prefix sums really help here—you can quickly get submatrix sums without recalculating every time. 3. It’s a nice extension of 1D subarray problems to 2D, so knowing those basics pays off big time. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving

Day: 57/365 📌 LeetCode POTD: Special Positions in a Binary Matrix Easy Key takeaways/Learnings from this problem: 1. This one shows how precomputing row and column counts makes checking each cell super easy. 2. Instead of re-scanning the whole row and column every time, count once and reuse smartly. 3. Good reminder that brute force can often be optimized with a tiny bit of preprocessing. 4. Clean implementation and careful condition checking matter more than complex algorithms here. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 74/365 📌 LeetCode POTD: Flip Square Submatrix Vertically Easy Key takeaways/Learnings from this problem: 1. This one highlights how matrix manipulation is mostly about correct indexing, not complex logic. 2. Flipping vertically is just swapping rows within the selected submatrix—visualizing it helps a lot. 3. Good reminder to be careful with boundaries when working on submatrices, not the whole grid. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

🚀 Day 13 of #LeetCode Challenge 🔢 Problem: 1758 – Minimum Changes To Make Alternating Binary String Today’s problem was about converting a binary string into an alternating pattern with minimum changes. 💡 Key Insight: An alternating string can only start with either ‘0’ or ‘1’. So instead of trying many possibilities, I compared both patterns and counted mismatches — then returned the minimum. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ This problem improved my thinking in: Pattern observation Optimization approach Writing clean and efficient Java code Consistency > Motivation 💪 #Day13 #Java #DSA #LeetCode #CodingJourney #FutureFullStackDeveloper

🚀 Day 20/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 350. Intersection of Two Arrays II Used sorting + two-pointer technique to efficiently find common elements appearing in both arrays while maintaining correct frequency. ⏱️ Time Complexity: O(n log n + m log m) 📦 Space Complexity: O(min(n, m)) (for storing the intersection) Strengthening understanding of array traversal and two-pointer pattern through consistent practice. 💪 Consistency keeps the progress moving 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 14/30 – LeetCode streak Today’s problem: Concatenation of Consecutive Binary Numbers  You need the decimal value of '1' + '2' + '3' + ... + 'n' written in binary back-to-back, all under mod (10^9 + 7). Core trick: treat “concatenate in binary” as shift + OR: * When you append 'i' to the right, you’re really shifting the current result left by 'bits(i)' and OR-ing 'i' into the free space. * The number of bits only increases when 'i' hits a power of two (1, 2, 4, 8, …), so you just track 'bits' and bump it whenever '(i & (i - 1)) == 0'. Day 14 takeaway: Once you see that “stick this binary to the right” is the same as “shift by its bit length and OR”, the whole problem becomes a clean for-loop plus the power-of-two trick—no string building or big integer juggling needed. #leetcode #dsa #java #bitmanipulation #consistency

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