LeetCode Solution: Array Concatenation in Java

🚀 Mastering Java Through LeetCode 🧠 Day 33 Not all nodes are “Good”… but finding them made me better at Trees 📌 LeetCode Problem Solved: Q.1448 – Count Good Nodes in Binary Tree 💭 Problem Summary: A node is called “Good” if no node on the path from root to it has a greater value. 🎯 Approach: ✔ Used DFS (Depth First Search) ✔ Tracked the maximum value seen so far in the path ✔ If current node ≥ maxSoFar → count it as a Good Node ✔ Updated max and continued recursion 🧠 Key Insight: Instead of storing the full path, just carry maxSoFar — simple yet powerful optimization! 💡 Complexity: ⏱ Time: O(N) 📦 Space: O(H) 🔥 What I Learned: How to maintain state during recursion Clean way to solve tree problems using DFS Thinking in terms of path-based conditions Consistency builds confidence 💪 Day 33 done… Let’s keep going 🚀 #Java #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #DSA #Tech #Learning

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Postorder Traversal (LeetCode 145) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform postorder traversal of the binary tree. • Recursively traverse the left subtree • Recursively traverse the right subtree • Visit the root node and store its value • Follow Postorder pattern: Left → Right → Root This approach works efficiently because recursion naturally follows the postorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the postorder pattern (Left → Right → Root), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Level Order Traversal (LeetCode 102) 💻 Language: Java 🔹 Approach: To solve this problem, I used Breadth-First Search (BFS) with a Queue to traverse the binary tree level by level. • Start by adding the root node into a queue • Use queue.size() to track the number of nodes at the current level • Process all nodes of that level using a loop • Store each level’s values in a separate list • Add left and right child nodes into the queue for the next level • Add each level list into the final result This approach efficiently groups nodes by levels and naturally fits the problem requirement. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) 📖 Key Learning: This problem strengthened my understanding of BFS, Queue operations, and level-wise tree traversal. It also reinforced how queue.size() helps process one complete level at a time — a very useful pattern for many tree problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #BFS #Queue

🧠 LeetCode POTD — The Biggest Hint Was Hidden in One Line 1855. Maximum Distance Between a Pair of Values Today's problem looked tricky at first considering the time complexity. We need to find indices (i, j) such that: 👉 i <= j 👉 nums1[i] <= nums2[j] And maximize: 👉 j - i ━━━━━━━━━━━━━━━━━━━ My first instinct? Try brute force combinations. Check many pairs. Maybe binary search. But I was missing the most important line in the question: Both arrays are non-increasing (sorted). That one detail changes everything. ━━━━━━━━━━━━━━━━━━━ 💡 Once I noticed that, the solution became a clean two-pointer approach. Start with: 👉 i = 0 in nums1 👉 j = 0 in nums2 Then: If nums1[i] <= nums2[j] 👉 This pair is valid 👉 Update answer with j - i 👉 Move j forward to try for a bigger distance Else: 👉 Current i is too large 👉 Move i forward Because arrays are sorted, once a pair fails, moving i is the only useful move. ━━━━━━━━━━━━━━━━━━━ 📌 What I liked about this problem: The challenge wasn't the coding. It was reading carefully enough to spot the property that unlocks the solution. ✅ Sometimes the answer is already in the problem statement. ✅ You just have to notice it. Curious — did anyone else miss the sorted hint at first? 👀 #LeetCode #TwoPointers #ProblemSolving #SoftwareEngineering #DSA #Java #c++ #SDE

🚀 DSA Every Day – Day 2 📍 Problem: Minimum Distance to Target Element 💡 Platform: LeetCode 🧠 Approach: Brute Force with Bidirectional Traversal 🔍 Problem Summary: Given an array, a target value, and a starting index, find the minimum distance between the start index and any index where the target exists. ⚙️ Approach Explained: Traverse forward (start → end) Traverse backward (start → beginning) For each occurrence of target: Compute distance → |start - i| Maintain the minimum distance 📈 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 🎯 Key Takeaways: Always think in terms of distance minimization problems Bidirectional traversal ensures complete coverage Can be optimized further using early stopping if needed Tell me your approach in comments ? Would love if people would like to join me in this habit for learning every day. 🔥 Progress: Day 2/ 60 #DSA #LeetCode #Java #CodingJourney #ProblemSolving #SoftwareEngineering #60DaysOfCode

📘 DSA Journey — Day 31 Today’s focus: Binary Search for boundaries and square roots. Problems solved: • Sqrt(x) (LeetCode 69) • Search Insert Position (LeetCode 35) Concepts used: • Binary Search • Search space reduction • Boundary conditions Key takeaway: In Sqrt(x), the goal is to find the integer square root. Using binary search, we search in the range [1, x] and check mid * mid against x to narrow down the answer. This avoids linear iteration and achieves O(log n) time. In Search Insert Position, we use binary search to find either: • The exact position of the target, or • The correct index where it should be inserted The key idea is that when the target is not found, the final position of the left pointer gives the correct insertion index. These problems highlight how binary search is not just for finding elements, but also for determining positions and boundaries efficiently. Continuing to strengthen fundamentals and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

🚀 Day 11 of Java with DSA Journey — This one blew my mind 🤯 📌 Problem: Power of Three (LeetCode 326) Yesterday, I used bit manipulation for Power of Two. Today? 👉 That approach completely fails. So I had to think differently… 💡 Breakthrough Idea Instead of loops or recursion… I used math + number theory to solve it in O(1) time. 👉 1162261467 % n == 0 Yes, one line. 🧠 Key Learnings 🔹 Bitwise isn’t universal Works great for base-2, but not for base-3 🔹 Prime Numbers Matter Since 3 is prime, its powers divide each other perfectly 🔹 Max Value Trick Largest power of 3 in 32-bit int = 3^19 = 1162261467 ⚡ Complexity ⏱ Time: O(1) 📦 Space: O(1) 🔥 Pro Tips (Interview Level) 💡 Tip 1: Know Your Data Type Limits Many O(1) tricks depend on constraints like 32-bit integer limits 💡 Tip 2: Prime Numbers Unlock Shortcuts If a number is prime, its powers have clean divisibility properties 💡 Tip 3: Always Question the Default Approach Most people write: while (n % 3 == 0) n /= 3; 💡 Tip 4: This is a Pattern Power of 2 → bitwise Power of 3 → math Power of 4 → hybrid 💡 Tip 5: Edge Cases First Always check n <= 0 🔥 Real Insight This problem forced me to shift my thinking: ❌ Rely on one technique ✅ Adapt based on the problem Consistency compounds 📈 #DSA #LeetCode #Java #CodingJourney #ProblemSolving #NumberTheory #InterviewPrep #Day11 #BitManipulation #CleanCode #Array #Optimization #MCA #lnct #100DaysOfCode #SoftwareEngineering #Algorithms #InPlaceAlgorithms #TechLearning #JavaDeveloper

🚀 Day 47of my #100DaysOfCode Journey Today, I solved the LeetCode problem: Mirror Distance of an Integer Problem Insight: Given a number, the task is to find the absolute difference between the original number and its reversed form. Approach: • Stored the original number in a temporary variable • Reversed the number using digit extraction (modulo and division) • Calculated the absolute difference between the original and reversed number Time Complexity: O(d), where d = number of digits Space Complexity: O(1) Key Learnings: • Digit manipulation using modulo and division is a powerful technique • Always store the original value before modifying the input • Reversing numbers is a fundamental pattern in many DSA problems Takeaway: Breaking the problem into simple steps makes even tricky-looking logic easy to solve. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Inorder Traversal (LeetCode 94) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform inorder traversal of the binary tree. • Recursively traverse the left subtree • Visit the root node and store its value • Recursively traverse the right subtree • Follow Inorder pattern: Left → Root → Right This approach works efficiently because recursion naturally follows the inorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the inorder pattern (Left → Root → Right), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Preorder Traversal (LeetCode 144) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform preorder traversal of the binary tree. • Visit the root node first • Store the root value in the result list • Recursively traverse the left subtree • Recursively traverse the right subtree • Follow Preorder pattern: Root → Left → Right This approach works efficiently because recursion naturally follows the preorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the preorder pattern (Root → Left → Right), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal