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In this video, we solve LeetCode Problem 1975 – Maximum Matrix Sum, a powerful greedy + mathematics based problem that often appears in technical interviews and competitive programming contests. 📘 Problem Overview You are given an n × n integer matrix. You can perform the following operation any number of times: 👉 Choose any two adjacent cells (sharing a border) and multiply both by -1. Your task is to maximize the sum of all elements in the matrix. 💡 Key Observation (Core Idea) Multiplying two adjacent elements by -1 does not change the parity of negative numbers. Therefore, we can rearrange signs freely except for one restriction: If the number of negative elements is odd, one element must remain negative. To maximize the sum: Make all elements positive If one must remain negative, choose the element with minimum absolute value 🧠 Optimal Strategy Convert all elements to their absolute values Calculate: Total sum of absolute values Count of negative elements Minimum absolute value in the matrix Apply logic: If negative count is even → return total sum If negative count is odd → subtract 2 × minimum absolute value ✨ Final Formula If negative_count is even: answer = sum(abs(matrix[i][j])) Else: answer = sum(abs(matrix[i][j])) - 2 × min_abs_value ⏱️ Complexity Analysis Time Complexity: O(n²) Space Complexity: O(1) (excluding input) 👨💻 Implementation Details Language used: C++ Efficient and clean logic Fully optimized for large constraints (n ≤ 250) Interview-ready approach 🎯 Who Should Watch This? ✔️ Students preparing for coding interviews ✔️ Competitive programmers ✔️ Anyone learning greedy algorithms ✔️ LeetCode practice enthusiasts #LeetCode1975 #MaximumMatrixSum #LeetCodeSolution #GreedyAlgorithm #MatrixProblems #DSA #CPlusPlus #CodingInterview #FAANGPreparation #CompetitiveProgramming #AlgorithmExplanation #MathInProgramming #ProblemSolving #InterviewQuestions #CodeAndMathematics