Intersection of Two Arrays II with Duplicates

🚀 Day 564 of #750DaysOfCode 🚀 🔍 LeetCode 3488 – Closest Equal Element Queries (Medium) Today’s problem was a really interesting mix of hashing + binary search + circular array logic — exactly the kind of question that tests both intuition and optimization skills. 💡 Problem Summary: Given a circular array nums and some query indices, for each query we need to find the minimum circular distance to another index having the same value. 🧠 Key Insight: Instead of checking every index (which would be too slow ❌), we: Store all indices of each number using a HashMap Use binary search to quickly find the closest neighbors Handle circular distance using: 👉 min(|i - j|, n - |i - j|) ⚡ Optimized Approach: Preprocess indices → O(n) Each query → O(log n) Overall → Efficient for large constraints (1e5) 📌 What I Learned: Circular arrays often require thinking in modulo arithmetic Preprocessing + binary search can drastically reduce complexity Always check edge cases (single occurrence) 🔥 Problems like this remind me that optimization is not about writing more code, but about thinking smarter. #LeetCode #Java #DSA #CodingJourney #ProblemSolving #HashMap #BinarySearch #Tech #SoftwareEngineering #LearnInPublic #Consistency

🚀 Day 76 — Slow & Fast Pointer (Find the Duplicate Number) Continuing the cycle detection pattern — today I applied slow‑fast pointers to an array problem where the values act as pointers to indices. 📌 Problem Solved: - LeetCode 287 – Find the Duplicate Number 🧠 Key Learnings: 1️⃣ The Problem Twist  Given an array of length `n+1` containing integers from `1` to `n` (inclusive), with one duplicate. We must find the duplicate without modifying the array and using only O(1) extra space. 2️⃣ Why Slow‑Fast Pointer Works Here  - Treat the array as a linked list where `i` points to `nums[i]`.   - Because there’s a duplicate, two different indices point to the same value → a cycle exists in this implicit linked list.   - The duplicate number is exactly the entry point of the cycle (same logic as LeetCode 142).  3️⃣ The Algorithm in Steps - Phase 1 (detect cycle): `slow = nums[slow]`, `fast = nums[nums[fast]]`. Wait for them to meet.   - Phase 2 (find cycle start): Reset `slow = 0`, then move both one step at a time until they meet again. The meeting point is the duplicate. 4️⃣ Why Not Use Sorting or Hashing?   - Sorting modifies the array (not allowed).   - Hashing uses O(n) space (not allowed).   - Slow‑fast pointer runs in O(n) time and O(1) space — perfect for the constraints. 💡 Takeaway:  This problem beautifully demonstrates how the slow‑fast pattern transcends linked lists. Any structure where you can define a “next” function (here: `next(i) = nums[i]`) can be analyzed for cycles. Recognizing this abstraction is a superpower. No guilt about past breaks — just another pattern mastered, one day at a time. #DSA #SlowFastPointer #CycleDetection #FindDuplicateNumber #LeetCode #CodingJourney #Revision #Java #ProblemSolving #Consistency #GrowthMindset #TechCommunity #LearningInPublic

🚀 Day 15 of LeetCode Problem Solving Solved today’s problem — LeetCode #49: Group Anagrams 💻🔥 ✅ Approach: HashMap + Sorting ⚡ Time Complexity: O(n * k log k) 📊 Space Complexity: O(n * k) The task was to group strings that are anagrams of each other. 👉 I used a HashMap where: Key = sorted version of string Value = list of anagrams 💡 Key Idea: If two strings are anagrams, their sorted form will be the same. 👉 Core Logic: Convert string → char array Sort the array Use sorted string as key Store original string in map 💡 Key Learning: Transforming data (like sorting strings) can help in identifying patterns and grouping efficiently. Consistency is the key — learning something new every day 🚀 #Day15 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

Day 6 of #100DaysOfLeetCode ✅ Today I solved LeetCode 128 — Longest Consecutive Sequence. 🧩 Problem Summary: Given an unsorted array of integers, the task is to find the length of the longest sequence of consecutive numbers. The challenge is to solve it in O(n) time complexity, which means sorting is not allowed. 💡 Key Learning: Instead of sorting, I used a HashSet for O(1) lookup. The main intuition: 👉 A number starts a sequence only if (num - 1) does NOT exist in the set. Then we expand forward (num + 1) to count the sequence length. ⚡ Concepts Practiced: • HashSet / Hashing • Optimized Searching (O(1) lookup) • Sequence Detection Pattern • Time Complexity Optimization 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) Every day improving problem-solving skills and understanding data structures deeper 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 100 Days of Code Day-25 LeetCode Problem Solved: Reverse Nodes in k-Group I recently worked on a Linked List problem that focuses on reversing nodes in groups of size k while preserving the remaining structure if the group size is less than k. 🔹 Strengthened my understanding of pointer manipulation 🔹 Improved problem decomposition skills 🔹 Practiced recursive thinking for efficient implementation 💡 Key takeaway: Breaking complex problems into smaller, manageable parts significantly simplifies the solution approach. Continuing to build consistency in problem-solving and deepen my understanding of Data Structures & Algorithms. #LeetCode #DataStructures #Algorithms #Java #ProblemSolving #SoftwareDevelopment

🚀 LeetCode Challenge 28/50 💡 Approach: Bidirectional HashMap + HashSet This is the word-level version of Isomorphic Strings! We need a true bijection — each pattern letter maps to exactly one word AND each word maps to exactly one letter. One map alone isn't enough! 🔍 Key Insight: → Split string s into words array → If lengths differ → instantly return false → Map char → word using HashMap → Track already-used words with a HashSet → If a new char tries to claim an already-used word → false! 📈 Complexity: ✅ Time: O(n) — single pass through pattern ✅ Space: O(n) — map + set storing at most n entries Pattern recognition in problem solving is just like pattern recognition in code — once you see the structure, the solution follows naturally! 🔍 #LeetCode #DSA #HashMap #Java #ADA #PBL2 #LeetCodeChallenge #Day28of50 #CodingJourney #ComputerEngineering #AlgorithmDesign #WordPattern

Day 2 of my LeetCode Journey Today’s focus: Prefix Sum + HashMap - Problem Type: Counting “Good Subarrays” - Approach: Converted the problem into a prefix sum transformation Used a HashMap to track frequencies Optimized brute force (O(n²)) → O(n) Key Learning: The real trick isn’t coding — it’s recognizing how to transform the problem. Instead of checking every subarray, I tracked a derived value: prefix - (index + 1) and counted how many times it appeared before. That shift turns an impossible brute force into a clean linear solution. -Takeaway: Most DSA problems are not about new algorithms — they’re about seeing the hidden pattern faster. Building consistency, one day at a time. #LeetCode #Day2 #DSA #Java #ProblemSolving #Consistency #CodingJourney

𝐃𝐚𝐲 𝟕𝟑 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on detecting duplicates within a given index range. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Contains Duplicate II --- 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 – 𝐇𝐚𝐬𝐡𝐌𝐚𝐩 • Stored each number with its latest index • For every element: • Checked if it already exists in the map • If yes → verified index difference ≤ k • Updated index to keep the most recent occurrence --- 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • Storing indices helps solve range-based problems • HashMap enables O(1) lookup • Updating latest index ensures correctness • Small constraints can change the approach --- 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(n) • Space: O(n) --- 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 Adding constraints doesn’t always make problems harder — it often makes them more interesting to solve. --- 73 days consistent 🚀 On to Day 74. #DSA #Arrays #HashMap #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

Day 68/100 | #100DaysOfDSA 🧩🚀 Today’s problem: Add Two Numbers A fundamental linked list problem that mimics real-life addition. Problem idea: Add two numbers represented by linked lists (digits stored in reverse order). Key idea: Linked list traversal + carry handling. Why? • We process digits one by one (like manual addition) • Need to handle carry at each step • Lists can have different lengths How it works: • Use a dummy node to build the result • Traverse both lists simultaneously • Add values + carry • Create new node with (sum % 10) • Update carry = sum / 10 • Move pointers forward • Continue until both lists and carry are done Time Complexity: O(max(m, n)) Space Complexity: O(max(m, n)) Big takeaway: Using a dummy node simplifies linked list construction and avoids edge cases. This pattern is very common in linked list problems. 🔥 Day 68 done. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #LinkedList #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity