Simple solution of MAXIMUM MATRIX SUM. (BEATS 100%)
 # Intuition  \n\n\n# Approach\n\n# CASE - 1\nIf the matrix has Zero or even numbers (n%2 = 0) of -ve elements.\n1. Just simply add the absolute values of each 
<!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n# **CASE - 1**\nIf the matrix has Zero or even numbers `(n%2 = 0)` of -ve elements.\n1. Just simply add the absolute values of each elements and store it in a variable (For eg. `ANS`).\n2. Return the `ANS`.\n\n# **CASE - 2**\nIf the matrix has odd numbers `(n%2 != 0)` of -ve elements.\n1. Add the absolute value of each elements and store it in a variable (For eg. `ANS`).\n2. Now find the minimum element and store it\'s absolute value (For eg. `min_element`).\n3. Return `Ans - (2*min_element)`.\n \n\n\n# Complexity\n- Time complexity: $$O(n2)$$\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    long long maxMatrixSum(vector<vector<int>>& matrix) {\n        long long count = 0, mini = INT_MAX, ans = 0;\n        int  n = matrix[0].size();\n        for (int i = 0; i < n; i++) {\n            for (int j = 0; j < n; j++) {\n                ans += abs(matrix[i][j]);\n                if (matrix[i][j] < 0) {\n                    count++;\n                }\n                if (abs(matrix[i][j]) < mini) {\n                    mini = abs(matrix[i][j]);\n                }\n            }\n        }\n         \n        if (count % 2 == 0)\n            return ans;\n        return ans - (2*mini);\n    }\n};\n```
['Array', 'Greedy', 'Matrix', 'C++']