LeetCode 53 | Maximum Subarray | Brute Force O(n²) to Kadane’s Algorithm O(n) | Java

#LeetCode53 #MaximumSubarray #KadanesAlgorithm #LeetCode #DSA LeetCode 53 – Maximum Subarray explained step by step in Java, starting with the Brute Force O(n²) approach and optimizing it to an efficient O(n) solution using Kadane’s Algorithm. In this video, we solve LeetCode 53: Maximum Subarray and understand how to move from a simple brute-force solution to the optimal Kadane’s Algorithm. If you are searching for LeetCode 53 solution, Maximum Subarray Java solution, Kadane’s Algorithm explanation, Maximum Subarray Sum, brute force to optimal solution, Blind 75 array problems, or coding interview preparation, this video covers the complete problem-solving process. We first understand the problem statement and examples, then solve the problem using a Brute Force O(n²) approach. After that, we identify the key observation that helps us optimize the solution to O(n) using Kadane’s Algorithm. The main idea behind Kadane’s Algorithm is simple: At every index, we decide whether: ✅ Start a new subarray from the current element OR ✅ Continue the previous subarray by adding the current element Core logic: currentSum = Math.max(nums[i], currentSum + nums[i]); maxSum = Math.max(maxSum, currentSum); This helps us find the maximum subarray sum in a single traversal. Example: nums = [-2,1,-3,4,-1,2,1,-5,4] Maximum Subarray = [4,-1,2,1] Maximum Sum = 6 ━━━━━━━━━━━━━━━━━━━━ 📌 In this video, you will learn: ✅ LeetCode 53 – Maximum Subarray ✅ Maximum Subarray Sum ✅ Brute Force O(n²) approach ✅ Kadane’s Algorithm ✅ Optimal O(n) solution ✅ Array traversal ✅ Continuous subarray concept ✅ Java solution ✅ Time and space complexity ✅ How to optimize brute-force solutions ✅ DSA problem-solving approach ✅ Blind 75 interview preparation ✅ Coding interview questions ✅ Campus placement preparation ━━━━━━━━━━━━━━━━━━━━ 📌 Problem Details Problem: LeetCode 53 – Maximum Subarray Difficulty: Medium Topic: Array, Dynamic Programming Language: Java Brute Force: Time Complexity: O(n²) Space Complexity: O(1) Optimal – Kadane’s Algorithm: Time Complexity: O(n) Space Complexity: O(1) ━━━━━━━━━━━━━━━━━━━━ 📌 Chapters 00:00 Problem Statement & Example 04:25 Brute Force Approach O(n²) 07:01 Brute Force Dry Run 08:50 Brute Force Java Solution in LeetCode 11:20 Kadane’s Algorithm Intuition 13:25 Kadane’s Algorithm Dry Run 16:50 Time & Space Complexity 17:01 Optimal O(n) Java Solution in LeetCode ━━━━━━━━━━━━━━━━━━━━ 🎯 Who should watch this video? This video is useful for: ✅ B.Tech / BE students ✅ Campus placement preparation ✅ Software engineering interviews ✅ DSA beginners ✅ Java developers ✅ LeetCode learners ✅ Blind 75 preparation ✅ Technical interview preparation ✅ Coding round preparation ✅ FAANG / product-based company interview preparation Maximum Subarray is one of the most important array problems because it teaches an important optimization pattern: Brute Force O(n²) → Optimal O(n) Understanding why we discard a negative running sum and start a new subarray is the key intuition behind Kadane’s Algorithm. If you are preparing for coding interviews, try solving the problem yourself before watching the optimal solution. Keep Learning, Keep Growing 🚀 👍 Like the video if it helped you 🔔 Subscribe for more LeetCode, Blind 75, DSA and Java interview problems 💬 Comment which LeetCode problem you want me to solve next #LeetCode53 #MaximumSubarray #KadanesAlgorithm #LeetCode #DSA #Java #Blind75 #CodingInterview #DataStructures #Algorithms #ProblemSolving #InterviewPreparation #PlacementPreparation