Optimizing Array Manipulation with In-Place Solution in Java

Day 83 of #100DaysOfCode Today’s problem: Ugly Number An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5. Approach: Instead of checking all factors, I kept dividing the number by 2, 3, and 5 repeatedly: If the number reduces to 1 → it’s ugly ✅ If something else remains → not ugly ❌ 🔍 Key Insight: Efficient problem-solving is often about reducing complexity, not increasing checks. 🧠 What I learned: How to simplify factor-based problems Importance of repeated division in optimization Writing clean and efficient logic ⚡ Example: 6 → 2 × 3 → Ugly ✅ 14 → includes 7 → Not Ugly ❌ Consistency is slowly building confidence 💪 On to Day 84! #DSA #Java #CodingJourney #ProblemSolving #LeetCode #100DaysOfCode

🚀 Day 6 of #100DaysOfCode | Pattern: Two Pointers Today I tackled a classic Linked List problem: 👉 Remove Nth Node From End of List At first, I thought of a brute force approach: Traverse the list to find length Then remove (length - n) node ❌ Works, but requires 2 passes 💡 Then I learned the optimal approach (Two Pointers): Use fast and slow pointers Move fast n steps ahead Move both together until fast reaches end Delete the node using slow ✅ Only 1 pass ✅ Cleaner and efficient ⚠️ Mistakes I made (important learning): Forgot to use a dummy node → failed for head deletion Got confused between moving n vs n+1 steps Faced null pointer issues in edge cases 🔥 Key Takeaway: Two pointer technique is not just for arrays, it’s super powerful in Linked Lists too 📈 Progress Update: Consistency building day by day Next week: Starting a new pattern Along with revision of Two Pointers #100DaysOfCode #DSA #Java #LinkedList #CodingJourney #ProblemSolving #LearnInPublic #PatternWise

🚀 Day 27/100 Days of Code Challenge Today’s problem: Find Peak Element (Leetcode 162) 🔍 What I learned: How to find a peak element efficiently without scanning the entire array Using Binary Search to reduce time complexity from O(n) to O(log n) Understanding how the “slope” of elements helps decide the search direction 🧠 Key Idea: Instead of checking every element, compare the middle element with its neighbor: If nums[mid] < nums[mid + 1] → move right Else → move left ✅ Example: Input: [1, 2, 3, 1] Output: 2 (index of peak element 3) Consistency is key 🔑 — improving problem-solving skills one day at a time! 💪 #Day27 #100DaysOfCode #LeetCode #BinarySearch #DSA #Java #CodingJourney

#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

I used to run parallel tasks using ExecutorService… But something always felt incomplete 🤔 Yes, tasks were running concurrently… But how do you know when ALL of them are done? That’s when I discovered the combo: 👉 ExecutorService + CountDownLatch And honestly, this is where multithreading started to make real sense. ⸻ 💡 The idea is simple: • ExecutorService → runs tasks in parallel ⚡ • CountDownLatch → makes sure you wait for all of them ⏳ ⸻ 🔥 Real flow: 1. Create a thread pool using ExecutorService 2. Initialize CountDownLatch with count = number of tasks 3. Submit tasks 4. Each task calls countDown() when done 5. Main thread calls await() 👉 Boom — main thread continues only when everything is finished ✅ ⸻ 🧠 Why this matters: ✔ Clean coordination between threads ✔ No messy shared variables ✔ Perfect for parallel API calls, batch processing, etc. ⸻ Before this, I was just “running threads” Now I’m actually controlling concurrency That’s a big difference. ⸻ If you’re learning backend or system design, this combo is 🔥 Simple tools… powerful impact. Have you used this pattern in real projects? 👇 #Java #Multithreading #ExecutorService #CountDownLatch #BackendDevelopment #LearningInPublic

Day 4/30 of #30DaysOfDSA 🔹 Problem: Top K Frequent Elements 💡 Approach: At first, I considered a straightforward approach — count the frequency of each element and then sort them based on frequency to pick the top k elements. While this works, the time complexity comes out to be O(n log n), which isn’t optimal for large inputs. To improve this, I explored a more efficient approach using Bucket Sort. I stored the frequency of each element using a HashMap, and instead of sorting, I grouped elements into buckets where the index represents their frequency. By iterating through these buckets from highest to lowest frequency, I was able to directly extract the top k frequent elements. This optimization reduced the time complexity to O(n), making the solution much more efficient. ⚡ What I Learned: • How to optimize solutions by eliminating unnecessary sorting • The power of HashMap for frequency counting • How Bucket Sort can be applied in real problem-solving scenarios #DSA #Coding #Consistency #Java

🚀 𝗗𝗮𝘆 𝟭𝟬 𝗼𝗳 𝟭𝟬𝟬: 𝗦𝗼𝗹𝘃𝗶𝗻𝗴 LeetCode! Today I focused on applying the 𝗣𝗿𝗲𝗳𝗶𝘅 𝗦𝘂𝗺 concept to identify balanced positions in an array. 📌 𝗣𝗿𝗼𝗯𝗹𝗲𝗺𝘀 𝗦𝗼𝗹𝘃𝗲𝗱 𝗧𝗼𝗱𝗮𝘆 (𝗗𝗮𝘆 𝟭𝟬) 🔹 𝗙𝗶𝗻𝗱 𝘁𝗵𝗲 𝗠𝗶𝗱𝗱𝗹𝗲 𝗜𝗻𝗱𝗲𝘅 𝗶𝗻 𝗔𝗿𝗿𝗮𝘆 ✅ Goal: Find an index where the sum of elements on the left is equal to the sum on the right. If multiple exist, return the leftmost one. Approach: Calculated the total sum of the array, then traversed it while maintaining a running left sum. At each index, compared left sum with "(total sum - left sum - current element)" to check if it satisfies the condition, achieving O(n) time complexity. #100DaysOfCode #DSA #LeetCode #Java #ProblemSolving #CodingJourney #PrefixSum #Arrays

💡 Day 39 of LeetCode Problem Solved! 🔧 🌟1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold 🌟 Task : Given an array of integers arr and two integers k and threshold, return the number of sub-arrays of size k and average greater than or equal to threshold.   Example 1: Input: arr = [2,2,2,2,5,5,5,8], k = 3, threshold = 4 Output: 3 Explanation: Sub-arrays [2,5,5],[5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other sub-arrays of size 3 have averages less than 4 (the threshold). Example 2: Input: arr = [11,13,17,23,29,31,7,5,2,3], k = 3, threshold = 5 Output: 6 Explanation: The first 6 sub-arrays of size 3 have averages greater than 5. Note that averages are not integers. #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

🔥 𝗗𝗮𝘆 𝟴𝟭/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟳𝟴. 𝗦𝘂𝗯𝘀𝗲𝘁𝘀 | 𝗠𝗲𝗱𝗶𝘂𝗺 | 𝗝𝗮𝘃𝗮 Given an integer array, return all possible subsets (the power set) — no duplicates allowed. 𝗠𝘆 𝗮𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗮𝗰𝗸𝘁𝗿𝗮𝗰𝗸𝗶𝗻𝗴: ✅ At each index, make a choice: include or exclude ✅ Recurse with the element added → then backtrack (remove it) ✅ Recurse again without the element ✅ Base case: when index == nums.length, save the current subset This explores every branch of the decision tree, giving us all 2ⁿ subsets cleanly. 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆: ⏱ Time: O(n × 2ⁿ) — 2ⁿ subsets, each copied in O(n) 📦 Space: O(n) recursion depth Backtracking is one of those patterns that feels magical once it clicks — the "add → recurse → remove" rhythm is the heart of so many classic problems. 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gZmezt_g 19 more days to go. Almost there! 💪 #LeetCode #Day81of100 #100DaysOfCode #Java #DSA #Backtracking #Recursion #CodingChallenge #Programming

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 🥷