Solved Missing Number in 100 Days of Code Challenge

LeetCode Challenge – Day 51 Today I solved the Diameter of Binary Tree problem. Problem Insight: The task is to find the length of the longest path between any two nodes in a binary tree. The path does not necessarily pass through the root. Approach: Used Depth First Search (DFS) to calculate the height of each subtree For every node: Calculated left subtree height Calculated right subtree height Updated the maximum diameter as the sum of left and right heights Returned the maximum diameter found during traversal Key Learning: Tree problems often combine multiple concepts. Here, calculating height and updating diameter together in a single traversal makes the solution efficient. Complexity: Time: O(n) Space: O(h) This problem helped me understand how recursion can be used to compute multiple values in a single traversal. #LeetCode #Java #DSA #CodingJourney

#100DaysOfCode – Day 2 Today I worked on a DSA problem based on arrays: Check if an array is sorted and rotated 🔍 Approach: Instead of finding the exact rotation point, I focused on identifying a pattern: In a sorted and rotated array, the order should break at most once. So, I checked how many times an element is greater than the next element while traversing the array in a circular manner. ✔️ If the count of such breaks is 0 or 1 → valid ❌ If it’s more than 1 → not a sorted rotated array 🧠 Key Takeaway: This problem taught me how pattern observation can simplify logic and avoid unnecessary complexity. Sometimes the best solution is not the most obvious one! 📈 Staying consistent and improving step by step 💪 #100DaysOfCode #DSA #DataStructures #Algorithms #Java #CodingJourney #ProblemSolving #LeetCode #Consistency

Day 19 of #100DaysOfCode Solved “Base 7” problem today — a simple yet insightful exercise on number systems. 💡 Approach: Convert the integer into base 7 by repeatedly dividing by 7 and storing remainders. Handle negative numbers separately and reverse the result at the end. ⚡ Key Learning: Even easy problems strengthen fundamentals like loops, math logic, and edge case handling. Consistency > Complexity. #LeetCode #DSA #CodingJourney #Java #ProblemSolving #100DaysOfCode

🚀 Day 66/100 Solved a problem on String Manipulation — Score of a String. 🔍 Problem Insight: Calculated the total score by summing absolute differences between ASCII values of adjacent characters. 💡 Approach: - Iterated through the string once - Used "Math.abs()" to compute differences - Maintained a running sum ⚡ Complexity: Time: O(n) Space: O(1) ✅ Result: All test cases passed Runtime: 1 ms (Beats 99.77%) 📚 Key Learning: Small optimizations and clean logic can lead to highly efficient solutions. Consistency is the key — showing up every day 💪 #Java #DSA #CodingJourney #LeetCode #ProblemSolving #100DaysOfCode

🚀 Day 31/200 – DSA Journey Solved: Contains Duplicate (LeetCode) 🔍 Approach Used: - Sorted the array - Compared adjacent elements - If nums[i] == nums[i-1], duplicate exists ⚡ Complexity: - Time: O(n log n) - Space: O(1) 💡 Key Learning: Brute force approach (O(n²)) is simple but not efficient. Using sorting helps optimize the solution and reduces time complexity. 📈 Progress Insight: Learning to move from brute force to optimized solutions step by step. Consistency > Motivation #DSA #LeetCode #Java #ProblemSolving #CodingJourney #SoftwareEngineering

Day 70 of My DSA Journey Problem: Linked List Cycle (LeetCode 141) Today’s problem was all about detecting whether a linked list contains a cycle — a classic and very important concept in data structures. Key Idea: Floyd’s Cycle Detection Algorithm (Tortoise & Hare) Instead of using extra memory, we use two pointers: • Slow pointer → moves 1 step • Fast pointer → moves 2 steps If there’s a cycle, these two pointers will eventually meet. If not, the fast pointer will reach the end (null). Insight: This approach is efficient because it avoids using extra space like a HashSet and still guarantees detection in linear time. Complexity: • Time: O(n) • Space: O(1) What I learned: Sometimes, the smartest solutions don’t require extra space — just a clever observation and pointer manipulation! Consistency > Perfection 💯 On to Day 71 🚀 #DSA #100DaysOfCode #Java #CodingJourney #ProblemSolving #LinkedList

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

📘 DSA Journey — Day 35 Today’s focus: Binary Search with index patterns. Problem solved: • Single Element in a Sorted Array (LeetCode 540) Concepts used: • Binary Search • Index parity (even/odd pattern) • Search space reduction Key takeaway: The array is sorted and every element appears twice except one. A key observation: Before the single element, pairs start at even indices After the single element, this pattern breaks. Using binary search: • If mid is even and nums[mid] == nums[mid + 1], the single element lies on the right side • Else, it lies on the left side (including mid) By leveraging this pattern, we can find the answer in O(log n) time and O(1) space. Continuing to strengthen binary search intuition and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

💻 Day 80 – DSA Practice (Binary Trees) Today’s challenge: 👉 Find all root-to-leaf paths where the sum equals a given target. 🔹 Approach: Use Depth-First Search (DFS) Track the current path and remaining sum When reaching a leaf node, check if the sum matches 💡 Key Concepts: Recursion + Backtracking Tree traversal (DFS) Efficient sum handling ⚡ Insight: Reducing the target sum at each step makes the solution cleaner and avoids recalculating sums. #DSA #BinaryTree #Java #CodingJourney #100DaysOfCode #ProblemSolving

Day 2/30 — DSA Challenge 🚀 Problem: Lowest Common Ancestor of Deepest Leaves Topic: Tree + DFS Difficulty: Medium Approach: Used DFS to calculate depth of left and right subtrees At each node, compared depths: If equal → current node is LCA Else → return deeper subtree result Mistake / Challenge: Initially overcomplicated thinking about storing all deepest nodes Realized we don’t need to track nodes explicitly, depth comparison is enough Fix: Returned a Pair (node, depth) from recursion Handled everything in one DFS traversal Key Learning: Tree problems often look complex but can be simplified with recursion Focus on what needs to be returned from each call Time Taken: 60 minutes Consistency check ✅ See you on Day 3. GitHub Repo: https://lnkd.in/g87geK_g #DSA #LeetCode #Java #Trees #Recursion #LearningInPublic