Linked List Random Node Challenge Solved in Java

🚀 Day 58 / 100 – LeetCode Challenge ✅ Problem Solved: Binary Tree Postorder Traversal 📊 Difficulty: Easy 💻 Language: Java Today’s problem focused on understanding tree traversal, specifically Postorder Traversal (Left → Right → Root). 🔍 Key Learnings: Mastered recursive approach for tree traversal Understood how DFS (Depth First Search) works in binary trees Strengthened problem-solving with clean and efficient logic ⚡ Performance: Runtime: 0 ms (Beats 100%) 🔥 Memory: 42.96 MB 💡 Approach: Used a simple recursive helper function: Traverse left subtree Traverse right subtree Add root node value This problem may be easy, but it builds a strong foundation for advanced tree problems 💪 📈 Consistency is key — 58 days down, 42 more to go! #LeetCode #100DaysOfCode #Java #DataStructures #CodingJourney #BinaryTree #DSA #Learning #Consistency

🔥 𝗗𝗮𝘆 𝟵𝟰/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟱𝟯𝟵. 𝗞𝘁𝗵 𝗠𝗶𝘀𝘀𝗶𝗻𝗴 𝗣𝗼𝘀𝗶𝘁𝗶𝘃𝗲 𝗡𝘂𝗺𝗯𝗲𝗿 | 🟢 Easy | Java Marked as Easy — but the optimal solution is pure binary search brilliance. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Given a sorted array, find the kth missing positive integer. Linear scan works — but can we do O(log n)? 💡 𝗧𝗵𝗲 𝗞𝗲𝘆 𝗜𝗻𝘀𝗶𝗴𝗵𝘁 At index i, the value arr[i] should be i+1 in a complete sequence. So missing numbers before arr[i] = arr[i] - 1 - i This lets us binary search on the count of missing numbers! ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 ✅ If arr[mid] - 1 - mid < k → not enough missing numbers yet, go right ✅ Else → too many missing, go left ✅ After the loop, left + k gives the exact answer 𝗪𝗵𝘆 𝗹𝗲𝗳𝘁 + 𝗸? After binary search, left is the index where the kth missing number falls beyond. left numbers exist in the array before that point, so the answer is left + k. No extra passes needed. ✨ 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(log n) — vs O(n) linear scan 📦 Space: O(1) This is a perfect example of binary searching on a derived condition, not just a value. A real upgrade from the naive approach. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gVYcjNS6 𝟲 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗧𝗵𝗲 𝗳𝗶𝗻𝗶𝘀𝗵 𝗹𝗶𝗻𝗲 𝗶𝘀 𝗿𝗶𝗴𝗵𝘁 𝘁𝗵𝗲𝗿𝗲! 🏁 #LeetCode #Day94of100 #100DaysOfCode #Java #DSA #BinarySearch #Arrays #CodingChallenge #Programming

Day 73 of #90DaysDSAChallenge Solved LeetCode 451: Sort Characters By Frequency Learned an important Java design concept today. Problem Overview: The task was to sort characters in a string based on descending frequency. What confused me initially: Why create a separate Freq class instead of just using HashMap and PriorityQueue directly? Key Learning: PriorityQueue stores one complete object at a time. For this problem, each item needs two pieces of data together: Character Frequency Example: Instead of storing: e and 2 separately We package them as: Freq('e', 2) That custom class acts like a container holding both values in one object, so PriorityQueue can compare and sort them correctly. Why this matters: This taught me that custom classes in Java are often not about complexity, they simply bundle related data into one manageable unit. Alternative approach: We can also use Map.Entry<Character, Integer> instead of creating a custom class, but building Freq makes the logic easier to understand while learning. Today’s takeaway: Not every class is for business logic — sometimes it exists just to package data cleanly. #Java #90DaysDSAChallenge #LeetCode #PriorityQueue #HashMap #CodingJourney #ProblemSolving

Day 74 of #100DaysOfLeetCode 💻✅ Solved #162. Find Peak Element problem in Java. Approach: • Used Binary Search technique to efficiently find the peak element • Set two pointers, left at start and right at end of the array • Calculated mid index using safe mid formula • Compared nums[mid] with nums[mid + 1] to determine direction • If mid element is smaller, moved search space to right half • Otherwise, moved search space to left half including mid • Continued until left and right pointers converged • Final position (left == right) represents the peak index Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 44.32 MB (Beats 25.49% submissions) Key Learning: ✓ Strengthened understanding of Binary Search on unsorted arrays ✓ Learned how to apply divide-and-conquer beyond traditional searching ✓ Improved intuition for peak finding using neighbor comparison ✓ Practiced optimizing search space instead of linear scanning Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Most people try to solve 3Sum using 3 nested loops. That solution works... but it's O(n³) and too slow. Today I learned how to reduce it to O(n²). 🚀 Day 81/365 — DSA Challenge Solved: 3Sum 🧠 The Idea Steps: 1. Sort the array 2. Fix one number 3. Use two pointers to find the other two numbers 4. Skip duplicates to avoid repeating triplets ⏱ Complexity Approach & Time 3 loops -> O(n³) Sort + Two pointers -> O(n²) Big improvement. 💡 What I learned today Whenever you see: • Sorted array • Pair sum • Triplet sum • Target sum Think: Sort + Two Pointers Day 81/365 complete. Still learning. Still coding. Code: https://lnkd.in/dad5sZfu #DSA #Java #LeetCode #LearningInPublic #100DaysOfCode #365Days365DSAProblems

Day 87 - LeetCode Journey Solved LeetCode 382: Linked List Random Node in Java ✅ This problem is all about picking a random node where every node has equal probability, even when the size of the linked list is unknown. Used the concept of Reservoir Sampling to solve this efficiently. While traversing the list, each node gets a fair chance to be selected without storing all values. No extra space, single pass, and perfectly balanced randomness. Key takeaways: • Learned Reservoir Sampling technique • Handling unknown-sized data efficiently • Achieving equal probability without extra space • Strengthening linked list traversal concepts ✅ All test cases passed ⚡ Optimal O(n) time, O(1) space solution This is one of those problems that feels simple but teaches a powerful concept 🔥 #LeetCode #DSA #Java #LinkedList #ReservoirSampling #Algorithms #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

Day 35 of #75DaysOfLeetCode 🚀 LeetCode 1448 — Count Good Nodes in Binary Tree Just solved an interesting tree problem that really strengthens DFS concepts! 🌳 👉 Problem Insight: A node in a binary tree is considered “good” if no node on the path from root to it has a value greater than it. 💡 Approach: Traverse the tree using DFS Keep track of the maximum value seen so far If current node ≥ max_so_far → it's a good node Update max and continue traversal 🧠 This problem is a great example of: ✔️ Tree traversal (DFS) ✔️ Passing state in recursion ✔️ Decision making at each node ⏱ Complexity: Time: O(n) Space: O(h) 💻 Java Code: class Solution { public int goodNodes(TreeNode root) { return dfs(root, root.val); } private int dfs(TreeNode node, int maxSoFar) { if (node == null) return 0; int count = 0; if (node.val >= maxSoFar) { count = 1; maxSoFar = node.val; } count += dfs(node.left, maxSoFar); count += dfs(node.right, maxSoFar); return count; } } 🔥 Key Takeaway: Always think about what information needs to be carried along the recursion path — here, it's the maximum value so far. #LeetCode #DataStructures #BinaryTree #DFS #Java #CodingInterview #100DaysOfCode

🚀 Mastering Java Through LeetCode 🧠 Day 32 Today I solved an Easy-level Tree problem that strengthened my understanding of Recursion + Depth Calculation — a fundamental concept for tree-based problems 📌 LeetCode Problem Solved: Q.104. Maximum Depth of Binary Tree 💭 Problem Summary: Given a binary tree, the task is to find the maximum depth — i.e., the number of nodes along the longest path from root to leaf. 🧠 Approach (Optimal): Instead of iterating level by level, I used a clean recursive approach: ✔️ If node is null → return 0 ✔️ Recursively calculate left subtree depth ✔️ Recursively calculate right subtree depth ✔️ Return 1 + max(left, right) ⚡ Key Learning: Recursion makes tree problems much simpler when you think in terms of “what should each function return for a node?” Complexity: Time: O(n) Space: O(h) Takeaway: Tree problems become easier once you master recursion and start recognizing patterns. Consistency is the key — showing up every day #Day32 #LeetCode #Java #DSA #CodingJourney #Recursion #BinaryTree #100DaysOfCode #SoftwareEngineering

Day 57 of #100DaysOfLeetCode 💻✅ Solved #506. Relative Ranks problem in Java. Approach: • Iterated through each player's score • Compared it with all other scores to determine its rank • Increased rank whenever a higher score was found • Assigned medals for top 3 ranks: Gold, Silver, Bronze • Converted remaining ranks to string and stored in result array Performance: ✓ Runtime: 124 ms (Beats 5.06% submissions) ✓ Memory: 46.76 MB (Beats 99.84% submissions) Key Learning: ✓ Understood ranking logic using comparison ✓ Practiced nested loop approach for problem solving ✓ Learned the importance of optimizing time complexity Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #Sorting #ProblemSolving #CodingJourney #100DaysOfCode

🚀 DAY 76/150 — CHECKING SUBTREES WITH RECURSION! 🌳 Day 76 of my 150 Days DSA Challenge in Java and today I solved a problem that strengthens my understanding of tree comparison and recursion 💻🧠 📌 Problem Solved: Subtree of Another Tree 📌 LeetCode: #572 📌 Difficulty: Easy–Medium The task is to determine whether one binary tree is a subtree of another binary tree. A subtree must match both structure and node values exactly. 🔹 Approach Used I used a recursive DFS approach: • Traverse each node of the main tree • For every node: Check if the subtree starting at that node is identical to the given tree • Use a helper function to compare two trees: If both nodes are null → true If values differ → false Recursively compare left and right subtrees ⏱ Complexity Time Complexity: O(n × m) (n = nodes in main tree, m = nodes in subtree) Space Complexity: O(h) (recursion stack) 🧠 What I Learned • Tree problems often require combining traversal + comparison • Recursion is powerful for handling hierarchical structures like trees • Breaking the problem into smaller checks simplifies the solution 💡 Key Takeaway This problem taught me how to: Compare two trees efficiently Apply recursion for structural matching Think in terms of subproblems within trees 🌱 Learning Insight As I continue my Binary Tree journey, I’m understanding that: Most tree problems are based on DFS, BFS, or recursion patterns Mastering these basics makes complex tree problems easier ✅ Day 76 completed 🚀 74 days to go 🔗 Java Solution on GitHub: 👉 https://lnkd.in/gZNXaW-C 💡 In trees, solving smaller parts correctly leads to the full solution. #DSAChallenge #Java #LeetCode #BinaryTree #Recursion #DFS #150DaysOfCode #ProblemSolving #CodingJourney #InterviewPrep #LearningInPublic