Solving 3354. Make Array Elements Equal to Zero with prefix sums and logic

Day 73/160 — Remove Loop in Linked List 🔁 After detecting a loop in a linked list, today’s challenge was to remove it safely — ensuring the list returns to its normal linear form. 💡 Approach: 1️⃣ Detect the loop using Floyd’s Cycle Detection Algorithm (Tortoise & Hare) 2️⃣ Find the starting node of the loop 3️⃣ Traverse to the last node of the loop and set its next pointer to null ⚙️ Complexity: O(n) time | O(1) space Breaking the cycle and restoring order — one pointer at a time! 💪 #Day73 #160DaysOfChallenge #DSA #Coding #LinkedList #Algorithms #Java #FloydCycleDetection #LearnToCode

📌 Day 12/100 – 100 Days of Code Challenge 🚀 Today’s LeetCode challenge: “Trapping Rain Water” (LeetCode #42) – one of the most popular Hard-level problems that strengthens logic building and array manipulation skills. 🔹 Approach: Used precomputed leftMax and rightMax arrays to determine how much water each bar can trap. For every index, the trapped water = min(leftMax[i], rightMax[i]) - height[i]. 🔹 Key Learnings: Improved understanding of prefix and suffix computations. Reinforced problem-solving using array-based preprocessing. Time Complexity: O(n) Space Complexity: O(n) #100DaysOfCode #LeetCode #Java #DSA #Arrays #TwoPointer #ProblemSolving #CodingJourney #DynamicProgramming #DailyChallenge

💻 Solving LeetCode: Kth Smallest Element in a BST Implemented an inorder traversal approach to efficiently find the kth smallest element in a Binary Search Tree. ✅ Key Points: Inorder traversal of BST gives elements in sorted order. Keep a count while traversing; when count == k, we’ve found our answer. Time Complexity: O(N) in worst case Space Complexity: O(H) due to recursion (H = height of tree) This is a classic example of leveraging BST properties for optimized search! #Coding #Java #DataStructures #BST #LeetCode #ProblemSolving #InorderTraversal

🚀 Day 40 of My #LeetCode Challenge 🚀 📘 Problem: Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold (LeetCode #1343) 💡 Approach: Implemented the Sliding Window technique to efficiently calculate the sum of each subarray of size k. First, compute the sum of the initial k elements. Then, slide the window forward — subtract the element leaving and add the new one entering. Check if the average of the current window meets or exceeds the given threshold. This method optimizes the time complexity to O(n) by avoiding repetitive summation. ⏱️ Concepts Used: Sliding Window | Arrays | Optimization | Time Complexity 🔥 Consistency and logic go hand in hand — every day adds a new layer to my problem-solving skills! #Day40 #LeetCode #100DaysOfCode #CodingChallenge #Java #DSA #ProblemSolving #CodingJourney #TechLearning #CodeEveryday #SoftwareEngineering #ProgrammerLife #CodingCommunity #Developer #JavaProgramming #LearnToCode #TechGoals #CodeNewbie #Algorithm #DataStructures #CodingPractice

#Day13 2 – Add Two Numbers Solved this classic linked list problem by implementing an efficient algorithm to add two numbers represented as reverse-digit linked lists. 💡 Approach: Used a dummy node to simplify list construction, iterated through both lists while handling different lengths, and maintained a carry value for proper digit summation. The solution elegantly handles cases where lists have different sizes and final carry propagation. #LeetCode #Java #LinkedList #DataStructures #Algorithms #ProblemSolving #Coding #Tech #SoftwareEngineering #DailyCoding #Programming

🚀 Day 47 of My LeetCode Journey 🚀 Problem: Sum of Square Numbers Topic: Math / Two Pointers 🧠 Approach: To check if a number c can be expressed as the sum of squares of two integers (a² + b² = c), we use the two-pointer technique: Start a = 0 and b = √c Calculate sum = a² + b² If sum == c → ✅ return true If sum < c → increase a If sum > c → decrease b Repeat until a <= b. ⏱️ Complexity: Time: O(√c) Space: O(1) This problem beautifully combines mathematical logic with an efficient two-pointer approach — clean and elegant! 💡 #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #DSA

🔥 Day 68 of #100DaysDSAChallenge 📌 Topic: Stacks & String Manipulation ✅ Problem Solved on #LeetCode: 3174. Clear Digits (🟢 Easy) 💡 Key Learnings: • Strengthened understanding of stack-based string modification logic. • Learned how to efficiently remove characters preceding digits using stack operations. • Improved ability to handle character-by-character simulations in string problems. • Reinforced understanding of stack data structure for real-world text processing use cases. 🚀 Consistency Matters: Each day of practice sharpens logic, builds confidence, and strengthens problem-solving mindset. Keep learning, keep building — growth compounds daily! 💪 🏷️ #100DaysChallenge #Day68 #DSA #LeetCode #ProblemSolving #CodingChallenge #Java #Programming #Stacks #LogicBuilding #Consistency #LearningTogether #10kCoders #10000Coders

🔥 LeetCode Day 28 Challenge Problem Attempted: Find Minimum in Rotated Sorted Array 💡 Approach & Progress: Today, I focused on identifying the minimum element in a rotated sorted array — a classic binary search variation that relies on recognizing the inflection (pivot) point. The key observation is that in a rotated array, all elements to the left of the pivot are greater than the elements to the right of the pivot. Using binary search, I compared the mid element with the rightmost element to decide which side to continue searching: If nums[mid] > nums[right], the minimum lies to the right. Otherwise, it lies on the left (including mid). This approach efficiently reduces the search space and finds the minimum in O(log n) time without needing to sort or traverse the array fully. 🧠 Key Learnings: Learned how to locate the pivot or minimum element using binary search logic. Understood how comparisons between mid and right help identify sorted vs rotated segments. Reinforced the principle that binary search can be used beyond simple value matching — for conditions and structure detection. Improved understanding of edge cases like unrotated arrays or minimal rotations. #LeetCode #Day28 #BinarySearch #DSA #ProblemSolving #CodingChallenge #Java #Algorithm #100DaysOfCode #LeetCodeJourney #CodeEveryday #TechLearning #ProgrammerLife

📌 Day 159 of Coding - Make Array Elements Equal to Zero (LeetCode - Easy) 🎯 Goal: Given an integer array nums, find the number of valid starting positions and directions such that by repeatedly decrementing and reversing direction (based on the rules), all elements eventually become 0. 💡Approach & Debugging: First computed total sum using Arrays.stream(nums).sum(). Maintained two prefix sums : left and right. For each index i where nums[i] == 0: Incremented left and decremented right progressively. Checked whether the left and right sums balance according to the movement rules. Counted valid selections when the balance was perfect or off by just one. ✔️ Time Complexity: O(N) - single pass through the array. ✔️ Space Complexity: O(1) - constant extra space. #Day159 #LeetCode #Simulation #Arrays #PrefixSum #Java #ProblemSolving #DSA #CodingChallenge #100DaysOfCode #Algorithms #InterviewPrep #SoftwareEngineering #DataStructures #CodeNewbie #ProgrammersLife