Arrays vs ArrayLists: Performance and Flexibility Compared

Some of the hardest problems become manageable once you recognize a repeating pattern. 🚀 Day 105/365 — DSA Challenge Solved: Subarrays with K Different Integers Problem idea: We need to count subarrays that contain exactly k distinct integers. Efficient approach: Use the powerful trick: subarrays with exactly k distinct = subarrays with ≤ k distinct − subarrays with ≤ (k − 1) distinct Steps: 1. Use a sliding window with a hashmap to track frequency of elements 2. Expand window by moving right pointer 3. If distinct count exceeds k, shrink window from the left 4. Count valid subarrays ending at each index 5. Subtract results to get exact count This pattern converts a hard problem into a manageable one. ⏱ Time: O(n) 📦 Space: O(n) Day 105/365 complete. 💻 260 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #HashMap #LeetCode #LearningInPublic

Day 66 — LeetCode Progress (Java) Problem: Find the Difference of Two Arrays Required: Given two integer arrays, return: Elements present in nums1 but not in nums2 Elements present in nums2 but not in nums1 Idea: Use sets to remove duplicates and quickly check membership. Approach: Convert both arrays into sets Iterate over nums1 set: Add elements not present in nums2 set to result1 Iterate over nums2 set: Add elements not present in nums1 set to result2 Return both lists Time Complexity: O(n + m) Space Complexity: O(n + m) #LeetCode #DSA #Java #HashSet #Arrays #ProblemSolving #CodingJourney

🚀 Day 35 of my #100DaysOfCode Journey Today, I solved the LeetCode problem: Valid Anagram Problem Insight: Given two strings, check if one is an anagram of the other. Approach: • First, check if the strings have the same length; if not, return false • Convert both strings to character arrays • Sort both arrays • Compare the sorted arrays — if equal, the strings are anagrams Time Complexity: • O(n log n) — due to sorting the arrays Space Complexity: • O(n) — for the character arrays Key Learnings: • Sorting is a simple and effective way to compare character compositions • Edge cases like different lengths should be handled first • Breaking the problem into small steps makes it easy to reason about Takeaway: Sometimes, sorting can reduce a seemingly complex problem into a simple comparison. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney #ProblemSolving #Strings

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

🚀 Day 7 – Exception Handling: More Than Just try-catch Today I focused on how exception handling should be used in real applications—not just syntax. try { int result = 10 / 0; } catch (Exception e) { System.out.println("Error occurred"); } This works… but is it the right approach? 🤔 👉 Catching generic "Exception" is usually a bad practice 💡 Better approach: ✔ Catch specific exceptions (like "ArithmeticException") ✔ Helps in debugging and handling issues more precisely ⚠️ Another insight: Avoid using exceptions for normal flow control Example: if (value != null) { value.process(); } 👉 is better than relying on exceptions 💡 Key takeaway: - Exceptions are for unexpected scenarios, not regular logic - Proper handling improves readability, debugging, and reliability Small changes here can make a big difference in production code. #Java #BackendDevelopment #ExceptionHandling #CleanCode #LearningInPublic

Day 77/100 Completed ✅ 🚀 Solved LeetCode – Search a 2D Matrix II (Java) ⚡ Implemented an efficient approach by starting from the top-right corner of the matrix and eliminating rows or columns based on comparison with the target. This reduces the search space at every step, achieving O(m + n) time complexity. 🧠 Key Learnings: • Smart traversal in a sorted 2D matrix • Eliminating search space using row & column properties • Moving left (col--) when value is greater • Moving down (row++) when value is smaller • Better than brute-force (O(m × n)) approach 💯 This problem improved my understanding of matrix traversal strategies and how to optimize searching using sorted properties. 🔗 Profile: https://lnkd.in/gaJmKdrA #leetcode #datastructures #algorithms #java #matrix #problemSolving #optimization #arrays #100DaysOfCode 🚀

🚀 Day 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

Taking the Two-Pointer technique to LeetCode! Today's challenge: Remove Duplicates from a Sorted Array. After practicing the Two-Pointer pattern for reversing arrays and swapping 0s and 1s, I applied it to a classic LeetCode problem. The challenge? Remove duplicates from an array in-place with O(1) extra memory. Because the array is already sorted, all duplicates are grouped together. • The Slow Pointer (The Writer): Keeps track of where the next unique element should be placed. • The Fast Pointer (The Reader): Scans ahead through the array to find new, unique numbers. • The Logic: If the Reader finds a duplicate, it just skips it. But the moment the Reader finds a new number, the Writer records it at the front of the array and steps forward. #DSA #Java #LeetCode #RemoveDuplicate

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 25/100: Finding the "Gap" 🎯 Today's challenge: Search Insert Position. We all know Binary Search finds an element in O(log n), but what if the element isn't there? I learned that by the end of the search, the `left` pointer doesn't just give up—it points exactly to where that missing number *should* be inserted to keep the array sorted. It’s a powerful way to handle dynamic data without breaking the order. Quarter of the way through the challenge! 🚀 #100DaysOfCode #Java #DSA #BinarySearch #ProblemSolving #Unit3 #Day25 #LearnInPublic