Solved Minimum in Rotated Sorted Array II with Binary Search

Day 43 of #100DaysOfLeetCode Today’s problem: Find Pivot Index Concepts practiced: Prefix Sum Technique Array Traversal Efficient Space Optimization Approach: The goal is to find an index such that the sum of all elements to its left equals the sum of all elements to its right. 1️⃣ First, calculate the total sum of the array. 2️⃣ Iterate through the array, and at each index: 🔹 Subtract the current element from the right sum. 🔹 Check if left sum equals right sum → if yes, that’s the pivot index! 3️⃣ Update the left sum and continue. #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #LearningInPublic #SoftwareEngineering #DSA

Just solved “Search in a Binary Search Tree” in Java with 100% runtime efficiency This problem was a great reminder of how elegant and efficient Binary Search Tree (BST) traversal can be when approached iteratively. 🔍 Key Idea: Start from the root and navigate left or right based on the target value — no recursion, just clean and optimized logic! 💡 Takeaway: Sometimes the simplest, most readable solutions are also the fastest. Writing clean and efficient code is a skill built through consistency and understanding the fundamentals. #Java #LeetCode #DSA #BinarySearchTree #CodingJourney #ProblemSolving #CleanCode

Day 44/100 – #100DaysOfCode 🚀 | #Java #LinkedList ✅ Problem Solved: Reverse Nodes in k-Group (LeetCode 25) 🧩 Problem Summary: Given a linked list, reverse every group of k consecutive nodes. If the last group has fewer than k nodes, leave it as is. 💡 Approach Used: We first count whether k nodes exist — only then we reverse that segment. Steps: Traverse ahead k nodes to ensure reversal is possible. Reverse exactly k nodes iteratively. Recursively process the remaining list. Connect the reversed part with the next processed segment. This allows efficient in-place operations without extra memory. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) (Reversing done in-place) ✨ Takeaway: This problem strengthens understanding of pointer manipulation and linked list segment handling — vital for clean and bug-free list operations. #Java #LinkedList #Pointers #Recursion #LeetCode #ProblemSolving #100DaysOfCode #CodingChallenge

🌳 Day 46 of #LeetCodeJourney Problem: 100. Same Tree ✅ Today’s challenge was about comparing two binary trees — checking if they’re structurally identical and have the same node values. A neat recursive problem that refreshes the basics of tree traversal and recursion! 💡 Key takeaway: If you can break a problem down into smaller identical subproblems — recursion will always have your back. #LeetCode #100DaysOfCode #Java #DSA #BinaryTree #Recursion #CodingJourney

Day 16: Solving “Find Pivot Index” – LeetCode (Java) Today, I explored an interesting problem called “Find Pivot Index” from LeetCode. It really helped me understand how to balance sums on both sides of an array element. Problem Statement: Given an array, find the index where the sum of all elements on the left is equal to the sum of all elements on the right. If no such index exists, return -1. Key Logic: For each element: Calculate leftSum = sum of elements before the index. Calculate rightSum = sum of elements after the index. Compare both. If they match → that’s your pivot index! 🔍 Example: Input: [1, 7, 3, 6, 5, 6] Output: 3 Explanation: Left sum = 11, Right sum = 11 Learning Moment: Concepts Used: Nested loops Prefix sums Edge case handling Logical debugging Next Step: I’ll now optimize this logic to a more efficient approach (using total sum & prefix sum) to make it run in O(n) time. #Java #LeetCode #DSA #ProblemSolving #LearningJourney #90DaysOfCode #BCA #CareerGrowth

Day 50 of #75DaysDSAChallenge Problem: 7. Reverse Integer Difficulty: 🟠 Medium Platform: LeetCode 🧩 Problem Statement Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes it to go outside the signed 32-bit integer range [-2³¹, 2³¹ - 1], return 0. Example: Input: x = 123 Output: 321 💡 Approach 1️⃣ Extract the last digit using x % 10. 2️⃣ Remove the last digit using x / 10. 3️⃣ Add the digit to the reversed number. 4️⃣ Before updating, check for overflow conditions. 5️⃣ Continue until all digits are processed. #LeetCode #Java #DSA #CodingChallenge #75DaysDSAChallenge #ProblemSolving #TechLearning #CodingJourney

Day 45/100 – #100DaysOfCode 🚀 | #Java #Hashing #SlidingWindow ✅ Problem Solved: Contains Duplicate III (LeetCode 220) 🧩 Problem Summary: Given an integer array and integers k and t, determine if there exist two distinct indices i and j such that: |i - j| ≤ k |nums[i] - nums[j]| ≤ t 💡 Approach Used: Used a Sliding Window + TreeSet to maintain numbers in a sorted structure. Steps: Traverse the array and maintain a window of at most size k. For each element x, find if there exists another element in the set such that |x - y| ≤ t. This is done using ceiling() to find the closest value ≥ x. Insert the element in TreeSet and remove the element that slides out. ⚙️ Time Complexity: O(n log k) 📦 Space Complexity: O(k) ✨ Takeaway: This problem teaches how ordered data structures like TreeSet help efficiently handle range queries in sliding window scenarios. #Java #TreeSet #SlidingWindow #LeetCode #ProblemSolving #100DaysOfCode #CodingChallenge

🔥 #100DaysOfDSA — Day 30/100 Topic: Reverse an Array in Java 🔁 💡 What I Did Today: Today, I explored how to reverse an array manually using the two-pointer approach. Instead of relying on built-in methods, I learned how to swap elements from both ends until the array is completely reversed. 🧠 Logic Used: Initialize two pointers: start = 0 (beginning of array) end = numbers.length - 1 (end of array) While start < end: Swap numbers[start] and numbers[end] Move pointers inward → start++ and end-- 📊 Example: Input → {2, 4, 6, 8, 10} Output → {10, 8, 6, 4, 2} ⚙️ Time Complexity: O(n) — each element is swapped once. O(1) space — done in-place without using extra arrays. ✨ Takeaway: Learning to reverse an array manually helps strengthen the concept of pointers, loops, and in-place operations. Sometimes the simplest logic gives the biggest "aha!" moment 💡 #100DaysOfCode #Day30 #Java #DSA #Arrays #ProblemSolving #CodingJourney #LearnInPublic #DeveloperLife #CodeNewbie #LogicBuilding

Day 39/100 – #100DaysOfCode 🚀 | #Java #LeetCode #HashMap ✅ Problem Solved: Minimum Index Sum of Two Lists 🧩 Problem Summary: Given two string arrays, find the common strings with the smallest index sum (index in list1 + index in list2). If multiple answers exist, return all of them. 💡 Approach Used: ➤ Stored the elements of the first list in a HashMap with their indices. ➤ Iterated the second list and checked for matches. ➤ Tracked the minimum index sum and collected all strings that match that sum. This avoids nested loops and improves efficiency. ⚙️ Time Complexity: O(n + m) 📦 Space Complexity: O(n) ✨ Takeaway: HashMaps continue to be one of the most effective tools for reducing time complexity through direct lookups 🔍⚡ #Java #LeetCode #HashMap #DataStructures #ProblemSolving #100DaysOfCode #CodingJourney

Day 29/100 – #100DaysOfCode 🚀 | #Java #LeetCode #BinaryTree ✅ Problem Solved: Verify Preorder Serialization of a Binary Tree 🌲 🧩 Problem Summary: Given a string representing a preorder serialization of a binary tree, determine if it’s valid. Example: "9,3,4,#,#,1,#,#,2,#,6,#,#" → ✅ valid "1,#" → ❌ invalid 💡 Approach Used: Used the slot-counting method 🧠 Each node uses one slot and non-null nodes create two new slots. Process the preorder string, ensuring slots never go negative and exactly zero remain at the end. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Takeaway: This problem teaches how binary tree structure can be validated with simple counting logic — no need to rebuild the tree! 🌱 #Java #LeetCode #BinaryTree #100DaysOfCode #ProblemSolving #CodingChallenge