Jayanthkumar Karthik’s Post

🚀 𝐒𝐨𝐥𝐯𝐞𝐝 𝐭𝐡𝐞 "𝐒𝐪𝐮𝐚𝐫𝐞𝐬 𝐨𝐟 𝐚 𝐒𝐨𝐫𝐭𝐞𝐝 𝐀𝐫𝐫𝐚𝐲" 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐨𝐧 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞! 🚀 The challenge was to take a sorted array of integers, square each number, and return a new array with these squares sorted in non-decreasing order. Instead of taking a naive approach with O(n log n) complexity, I implemented an efficient two-pointer technique to achieve an optimal O(n) solution. 𝐇𝐢𝐠𝐡𝐥𝐢𝐠𝐡𝐭𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: > Two-pointer approach: Start from both ends of the array, compare the squared values, and fill the output array from the end. > Optimal time complexity: Achieves O(n) performance, which is essential for handling large input sizes. > Space efficiency: Uses an additional array for the output but avoids unnecessary sorting operations. Check out my solution and let me know your thoughts! I'm always open to feedback and discussions about algorithm optimization. #LeetCode #Java #CodingChallenge #Algorithms #ProblemSolving #Tech #SoftwareEngineering #Optimization #Coding

🚀 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞: 𝐒𝐞𝐚𝐫𝐜𝐡 𝐚 𝟐𝐃 𝐌𝐚𝐭𝐫𝐢𝐱 🚀 I recently solved the Search a 2D Matrix problem from LeetCode’s Top Interview 150 list, which focused on efficiently finding a target value in a 2D matrix using binary search. By treating the matrix like a flattened array and applying binary search, the problem is reduced to an elegant O(log(m * n)) solution. 𝐊𝐞𝐲 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲𝐬: 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(log(m * n)) 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(1) #coding #leetcode #interviewpreparation #java #datastructures #binarysearch

🌟 Day 26 of #100DaysOfCode! Today, I dove into some matrix manipulation and tackled LeetCode Problem #48: Rotate Image. 🚀 Here’s what I accomplished: 🌀 Rotate Image: I implemented the algorithm to rotate a given 2D matrix by 90 degrees clockwise. This challenge helped me improve my understanding of matrix indexing and transformations. 💻 Key Takeaways: - Practiced in-place array manipulation. - Reinforced my knowledge of matrix traversal techniques. If you're also working on coding challenges or exploring algorithms, let’s connect and motivate each other! Keep pushing forward! 💪 #100DaysOfCode #LeetCode #MatrixManipulation #Java #DataStructures #CodingJourney #Learning #Developer

🚀Leetcode Problem For today I recently tackled a problem that required finding all elements appearing more than 〈n/3〉 times in an array. This challenge led me to utilize a combination of strategies, and I'm thrilled with the outcome! 🔍 How I Approached the Problem: The problem was centered around identifying elements that appeared more than 〈n/3〉 times in an array. This observation was key because it meant that there could be, at most, only two such elements. 💡 Algorithm Choice: I employed the Boyer-Moore Voting Algorithm, which is typically used for finding a majority element (one appearing more than 〈n/2〉 times). However, by adapting this algorithm, I was able to efficiently find the potential candidates and verify them. 📈 Results: Runtime: 2 ms ⚡ Memory: 48.30 MB 🧠 Performance: Beat 99.47% of submissions! 🏆 This solution not only enhanced my understanding of the Boyer-Moore Voting Algorithm but also reinforced the importance of adapting well-known algorithms to fit specific problem constraints. #Java #Algorithms #Coding #ProblemSolving #DataStructures #LeetCode #BoyerMoore

🚀 #100DaysOfLeetCodeChallenge - Day 63 🚀 Today's challenge was to tackle the "Remove Duplicates from Sorted Linked List" problem. A classic interview question, this task requires us to efficiently remove duplicate elements from a sorted linked list. 🔑 Approach: Traverse the linked list while comparing each node with the next one. If the value of the current node is the same as the next, skip the next node. This solution works in O(n) time complexity, making it efficient for large lists. 📈 Key Takeaways: Understanding how to manipulate pointers in linked lists is crucial for solving these types of problems. This solution runs in O(n) time, with O(1) space complexity, making it optimal for large datasets. Let's continue the grind! 💪 #LeetCode #Java #LinkedList #Algorithm #CodingChallenge #ProblemSolving #SoftwareEngineering #TechJourney

"Day 16 of Tackling Leetcode Problem Of the Day" 🚀 Exploring Tree Equivalence: Flip Equivalent Binary Trees 🌳💡 I recently tackled an interesting problem—determining if two binary trees are flip equivalent. This involves checking if two trees can become identical through a series of flip operations at any level. Here’s an elegant recursive solution that does the job efficiently! 🔍 Problem #951 Summary: Given two binary trees, we need to check if they are flip equivalent, meaning they are structurally identical or can become identical by flipping some of their children at different levels. 🔑 Key Concepts: • Recursion: The solution recursively compares nodes at each level. • Flip Check: We check if the children of both nodes are either in the same order or flipped. • Base Cases: Handle null trees and mismatched values upfront for early termination. 💡 Takeaway: This problem is a fantastic exploration of recursion in binary trees, where both symmetry and structural changes (flipping) are considered to determine tree equivalence. 💬 How do you approach recursive tree problems? Let’s discuss in the comments below! 👇 #BinaryTrees #TreeAlgorithms #Recursion #FlipEquivalence #LeetCode #Java #CodingSkills #EfficientCoding #ProblemSolving #Tech

𝗧𝗼𝗱𝗮𝘆'𝘀 𝗟𝗲𝗲𝘁𝗰𝗼𝗱𝗲 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻: 𝟭𝟱𝟱𝟬. 𝗧𝗵𝗿𝗲𝗲 𝗖𝗼𝗻𝘀𝗲𝗰𝘂𝘁𝗶𝘃𝗲 𝗢𝗱𝗱𝘀 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻: Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false. 𝗘𝘅𝗮𝗺𝗽𝗹𝗲 𝟭: 𝗜𝗻𝗽𝘂𝘁: arr = [2, 6, 4, 1] 𝗢𝘂𝘁𝗽𝘂𝘁: false 𝗘𝘅𝗽𝗹𝗮𝗻𝗮𝘁𝗶𝗼𝗻: There are no three consecutive odds. 𝗘𝘅𝗮𝗺𝗽𝗹𝗲 𝟮: 𝗜𝗻𝗽𝘂𝘁: arr = [1, 2, 34, 3, 4, 5, 7, 23, 12] 𝗢𝘂𝘁𝗽𝘂𝘁: true 𝗘𝘅𝗽𝗹𝗮𝗻𝗮𝘁𝗶𝗼𝗻: [5, 7, 23] are three consecutive odds. 𝗠𝘆 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵: 𝟭. 𝗜𝗻𝗶𝘁𝗶𝗮𝗹𝗶𝘇𝗲 𝗮 𝗖𝗼𝘂𝗻𝘁𝗲𝗿: Create a counter to keep track of consecutive odd numbers. 𝟮. 𝗜𝘁𝗲𝗿𝗮𝘁𝗲 𝗧𝗵𝗿𝗼𝘂𝗴𝗵 𝘁𝗵𝗲 𝗔𝗿𝗿𝗮𝘆: Traverse through each element of the array. 𝟯. 𝗖𝗵𝗲𝗰𝗸 𝗶𝗳 𝘁𝗵𝗲 𝗰𝘂𝗿𝗿𝗲𝗻𝘁 𝗲𝗹𝗲𝗺𝗲𝗻𝘁 𝗶𝘀 𝗼𝗱𝗱: If it is, increment the counter. If it is not, reset the counter to zero. 𝟰. 𝗖𝗵𝗲𝗰𝗸 𝗶𝗳 𝘁𝗵𝗲 𝗰𝗼𝘂𝗻𝘁𝗲𝗿 𝗿𝗲𝗮𝗰𝗵𝗲𝘀 𝘁𝗵𝗿𝗲𝗲: If it does, return true. 𝟱. 𝗥𝗲𝘁𝘂𝗿𝗻 𝘁𝗵𝗲 𝗥𝗲𝘀𝘂𝗹𝘁: After processing all elements, if the counter never reaches three, return false. 𝗧𝗶𝗺𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆: O(n) 𝗦𝗽𝗮𝗰𝗲 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆: O(1) #Leetcode #Coding #Programming #Java #Algorithms #DataStructures #Array #OddNumbers #ConsecutiveOdds

1. N-Queens Problem: I solved the N-Queens problem using backtracking. The idea is to place a queen in each row, ensuring no two queens can attack each other by checking column and diagonal conflicts. The approach explores all possible placements and backtracks when a conflict occurs. Time Complexity: O(n!) Space Complexity: O(n) . 2. Combination Sum II Problem: I approached the Combination Sum II problem with backtracking to find all unique combinations that sum up to a target. I used sorting to efficiently skip duplicates and pruned the search space when the current sum exceeded the target. Time Complexity: O(2^n) Space Complexity: O(n) . Mohammad F. LearnYard Technologies FZ-LLC #NQueensProblem #Backtracking #AlgorithmDesign #Coding #ProblemSolving #LeetCode #Tech #Java #DataStructures #Recursion #ComputerScience #SoftwareEngineering #frazsheet #dsasheet

🧠 Space Complexity: Memory Matters in Algorithm Design... 📚 What is Space Complexity? Space complexity measures the amount of memory an algorithm needs to run. It's essential for optimizing performance, especially when working with large datasets or resource-constrained systems. 🚀 Common Space Complexities: O(1) – Constant space O(n) – Linear space O(n²) – Quadratic space 🔍 Question for You: When optimizing algorithms, do you focus more on time or space complexity? Let’s discuss below! . . . . #softwareengineering #java #springboot #microservices #coding

✨ Day 57/60 of my #60DaysCodeChallenge ✨ Today, I tackled the Find Minimum in Rotated Sorted Array problem using binary search! 🔍 💡 Approach: Binary Search: Since the array is sorted and rotated, binary search helps locate the minimum element in O(log n) time. Logic: By comparing the middle element with the rightmost element, we can determine which half of the array is sorted and adjust our search accordingly. Efficiency: This approach is much faster than linearly scanning the array or sorting it first. It allows us to find the minimum element in large datasets efficiently. #Coding #BinarySearch #Java #CodingChallenge #TechJourney #SoftwareDevelopment #ProblemSolving #SimarpreetSingh