Maximizing Average Subarray with Sliding Window Technique

🚀 Day 25 – LeetCode Challenge 📌 Problem: Subarray Product Less Than K 💡 Approach: Sliding Window (Two Pointers) ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Today I solved a classic sliding window problem where we count all contiguous subarrays whose product is strictly less than k. 🔍 Key Insight: Since all elements are positive, we can maintain a running product and shrink the window whenever the product becomes ≥ k. This avoids the brute-force O(n²) solution and optimizes it to O(n). 🔥 Sliding Window pattern getting stronger day by day! #Day25 #LeetCode #CodingJourney #DSA #Java #PlacementPreparation

Day 45 on LeetCode Largest Submatrix With Rearrangements :- This problem clicked once I stopped treating columns as fixed. Rearranging them turns each row into a controllable histogram. Approach: ->Build vertical heights of consecutive 1s ->Sort each row to simulate optimal column order ->Evaluate max area using height × width Time Complexity: O(m × n log n) The editorial on LeetCode made the difference here. It didn’t just give the solution, it clarified the intuition behind sorting heights, which is easy to miss on your own. #DSA #LeetCode #Java

🚀 Day 89 of #100DaysofLeetCode Problem: Minimum Size Subarray Sum Difficulty: Medium Today’s challenge was about finding the smallest length subarray whose sum is greater than or equal to a given target. Since all elements are positive, this problem is a perfect fit for the Sliding Window technique. Instead of checking all possible subarrays (which would be inefficient), I used two pointers: Expand the window by moving the right pointer and adding to the current sum. Once the sum becomes ≥ target, shrink the window from the left to find the minimum possible length. Keep updating the minimum length during this process. Key takeaways: Positive integers allow efficient window shrinking. Two pointers help reduce time complexity from O(n²) to O(n). Maintain a running sum and dynamically adjust window size. Return 0 if no valid subarray exists. Time Complexity: O(n) Space Complexity: O(1) Another solid sliding window problem completed 💪 #100DaysOfCode #100DaysOfLeetCode #LeetCode #DSA #SlidingWindow #TwoPointers #Java #ProblemSolving #CodingJourney #Algorithms

LeetCode Problem || Count Submatrices with Top-Left Element and Sum Less Than k (3070) 🚀 🔍 Approach : I implemented a 2D Prefix Sum technique to efficiently calculate the sum of submatrices. Instead of recalculating sums again and again, I built a presum matrix where each cell stores the cumulative sum from the top-left corner. Key Idea: Use inclusion-exclusion principle: Current = Grid Value + Left + Top − Top-Left If the sum at any point exceeds k, break early to optimize performance. ⚡ Time Complexity: O(m × n) ⚡ Space Complexity: O(m × n) #LeetCode #DSA #Java #Coding #ProblemSolving #LearningJourney

LeetCode Problem || Find All Possible Stable Binary Arrays II (3130)🚀 🔹 Problem Idea: We are given a number of 0s and 1s and a constraint called limit. The goal is to count the number of possible binary arrays such that no more than limit consecutive identical elements appear. 🔹 Approach: I used Dynamic Programming to track the number of valid arrays formed with: i zeros j ones and the last placed element (0 or 1). The DP state helps ensure that the consecutive limit condition is maintained while building the array. Always enjoying the process of learning and improving problem-solving skills! 💡 #LeetCode #DynamicProgramming #ProblemSolving #Java #CodingPractice #SoftwareEngineering

🚀 Day 2 – LeetCode Challenge Today I solved the Longest Common Prefix problem. Day 2 of #50DaysLeetcodeChallenge 📌 Problem Statement Given an array of strings, we need to find the longest common prefix shared among all the strings. If no common prefix exists, we return an empty string "". Example: ["flower", "flow", "flight"] → "fl" ["dog", "racecar", "car"] → "" 💡 Approach I Used Assume the first string in the array as the prefix. Compare this prefix with the remaining strings one by one. If the current string does not start with the prefix, reduce the prefix by removing the last character. Repeat this until the string starts with the prefix. Continue the process for all strings. This gradually shrinks the prefix until it matches all strings. #LeetCode #Java #CodingChallenge #DSA #SoftwareEngineering #LearningInPublic

#Day32 of #365DaysOfCode Today’s LeetCode Practice: 🔹 Container With Most Water (LeetCode 11) Solved a classic two-pointer optimization problem where the goal is to find two lines that together form a container holding the maximum water. 💡 Key Insight: Start with two pointers at both ends. Calculate area using: width × min(height[left], height[right]) Move the pointer with the smaller height inward, because the smaller height limits the water capacity. This guarantees exploring better possibilities efficiently. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Consistency > Motivation. Day by day, improving problem-solving and logical thinking skills #LeetCode #ProblemSolving #Java #CodingJourney #FutureEngineer #Consistency #LearningEveryday

How to Get 0 ms on LeetCode 🚀😂 Solved a problem. Got 3 ms. Felt smart for a minute 🧠✨ Then I saw someone with 0 ms. Ego slightly hurt. Am I getting old? Did someone unlock O(0)? Is there a secret JVM setting I missed?? 😭 So I opened their solution… and found this: static { Runtime.getRuntime().addShutdownHook(new Thread(() -> { try (FileWriter fw = new FileWriter("display_runtime.txt")) { fw.write("000"); } })); } Bro didn’t optimize the code. Bro optimized the scoreboard 🏆😂 Honestly… I respect the creativity 🤝🔥 #LeetCode #Java #CodingLife

🚀 Day 14/100 – #100DaysOfCode Challenge 🔍 Problem Solved: Single Number (LeetCode 136) Today’s problem was about finding the element that appears only once in an array where every other element appears twice. 💡 Key Insight: Used the power of XOR (^) bit manipulation Same numbers cancel out → a ^ a = 0 XOR with 0 gives the number → a ^ 0 = a 👉 By XOR-ing all elements, duplicates vanish, leaving the unique number! ⚡ Approach: Traverse the array once Apply XOR on each element Result = single number ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) 🔥 What I Learned: Bit manipulation can simplify problems drastically XOR is super powerful for pairing problems #Day14 #CodingChallenge #Java #DataStructures #Algorithms #BitManipulation #LeetCode #100DaysOfCode #CodingJourney