Flatten Binary Tree to Linked List in-place with Preorder Traversal

Day 76/100 | #100DaysOfDSA 🧩🚀 Today’s problem: Lowest Common Ancestor of a Binary Tree A fundamental tree problem that builds strong recursion intuition. Problem idea: Find the lowest node in the tree that has both given nodes as descendants. Key idea: DFS + recursion (bottom-up approach). Why? • Each subtree can independently tell if it contains p or q • Combine results while backtracking • First node where both sides return non-null → answer How it works: • If current node is null / p / q → return it • Recursively search left and right subtree • If both left & right are non-null → current node is LCA • Else return the non-null side Time Complexity: O(n) Space Complexity: O(h) (recursion stack) Big takeaway: Tree problems often rely on post-order traversal + combining child results. Understanding this pattern unlocks many binary tree problems. 🔥 Day 76 done. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #BinaryTree #Recursion #DFS #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

**Day 116 of #365DaysOfLeetCode Challenge** Today’s problem: **Next Greater Element II (LeetCode 503)** A classic **Monotonic Stack** problem with a twist: 👉 The array is **circular** So after the last element, we continue from the first element. 💡 **Core Idea:** For every number, find the **first greater element** while traversing forward. If none exists → return `-1` Example: Input: `[1,2,1]` Output: `[2,-1,2]` Why? * First `1 → 2` * `2 → no greater` * Last `1 → wrap around → 2` 📌 **Efficient Approach: Monotonic Stack** Use stack to store indices whose next greater element is not found yet. Traverse array **twice**: `0 → 2*n - 1` Use: `idx = i % n` This simulates circular behavior. Whenever current number is greater than stack top element: 👉 Pop index 👉 Update answer ⚡ **Time Complexity:** O(n) ⚡ **Space Complexity:** O(n) **What I learned today:** Circular array problems often become simple when you traverse twice using modulo. 👉 `i % n` This trick appears in many advanced array questions. 💭 **Key Takeaway:** When you see: * Next Greater Element * Previous Smaller Element * Nearest Bigger Value #LeetCode #DSA #MonotonicStack #Stack #Arrays #Java #CodingChallenge #ProblemSolving #TechJourney #Consistency

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

Day 67/100 | #100DaysOfDSA 🧩🚀 Today’s problem: Subsets II Another classic backtracking problem with a twist (duplicates). Problem idea: Generate all possible subsets (power set), but avoid duplicate subsets. Key idea: Backtracking + sorting to handle duplicates. Why? • We need to explore all subset combinations • Duplicates in input can lead to duplicate subsets • Sorting helps us skip repeated elements efficiently How it works: • Sort the array first • At each step, add current subset to result • Iterate through elements • Skip duplicates using condition: 👉 if (i > start && nums[i] == nums[i-1]) continue • Choose → recurse → backtrack Time Complexity: O(2^n) Space Complexity: O(n) recursion depth Big takeaway: Handling duplicates in backtracking requires careful skipping logic, not extra data structures. This pattern appears in many problems (subsets, permutations, combinations). 🔥 Day 67 done. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #Backtracking #Recursion #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

🚀 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 59/100 | #100DaysOfDSA 🔍⚡ Today’s problem: Single Element in a Sorted Array A clever Binary Search problem with a twist. Key observation: In a sorted array where every element appears twice except one, pairs follow a pattern. Before the single element: • First occurrence → even index • Second occurrence → odd index After the single element: • Pattern breaks Approach: • Use binary search • Check mid with its pair using index trick (mid ^ 1) • If nums[mid] == nums[mid ^ 1] → move right • Else → move left This helps pinpoint where the pattern breaks. Time Complexity: O(log n) Space Complexity: O(1) Big takeaway: Bit manipulation + pattern observation can simplify binary search problems significantly. Small tricks → big optimizations. 🔥 Day 59 done. #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearch #BitManipulation #Arrays #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

Day 56/100 | #100DaysOfDSA 🧠⚡ Today’s problem: String Compression A clean in-place array manipulation problem. Core idea: Compress consecutive repeating characters and store the result in the same array. Approach: • Traverse the array using a pointer • Count consecutive occurrences of each character • Write the character to the array • If count > 1 → write its digits one by one • Move forward and repeat Key insight: We don’t need extra space — just carefully manage read & write pointers. Time Complexity: O(n) Space Complexity: O(1) Big takeaway: In-place algorithms require precise pointer control but give optimal space efficiency. Mastering these improves real-world memory optimization skills. 🔥 Day 56 done. #100DaysOfCode #LeetCode #DSA #Algorithms #Strings #TwoPointers #InPlace #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

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

Day 58/100 | #100DaysOfDSA 🔄🔍 Today’s problem: Search in Rotated Sorted Array A twist on Binary Search with a rotated array. Key idea: Even though the array is rotated, one half is always sorted. Approach: • Use binary search • Find mid element • Check which half is sorted (left or right) • Decide if target lies in the sorted half • Narrow the search accordingly Why it works: At every step, we eliminate half of the search space just like standard binary search. Time Complexity: O(log n) Space Complexity: O(1) Big takeaway: Understanding the structure of the problem helps adapt classic algorithms like binary search. Rotation doesn’t break order — it just shifts it. 🔥 Day 58 done. #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearch #Arrays #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

Day 57/100 | #100DaysOfDSA ⛰️🔍 Today’s problem: Peak Index in a Mountain Array A perfect use-case of Binary Search on a pattern. Key observation: The array increases → reaches a peak → then decreases. Approach: • Use binary search on the index • Compare mid with mid + 1 • If arr[mid] > arr[mid + 1] → we are on the decreasing side → move left • Else → we are on the increasing side → move right This guarantees we always move toward the peak. Time Complexity: O(log n) Space Complexity: O(1) Big takeaway: Binary search isn’t just for sorted arrays — it works on patterns too. Recognizing these patterns is a game changer. 🔥 Day 57 done. #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearch #Arrays #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity