Maximize Matrix Sum with Optimal Sign Flips
The provided Java code solves the problem of calculating the maximum possible sum of a matrix after any number of sign flips (changing positive to negative or v
The provided Java code solves the problem of calculating the **maximum possible sum of a matrix** after any number of sign flips (changing positive to negative or vice versa) on the elements of the matrix.\n\n### Key Insights and Conclusion:\n\n1. **Absolute Values and Negatives**:\n   - The algorithm processes the matrix to compute the sum of the absolute values of all elements (`sum`) and counts the number of negative elements (`negCount`).\n   - It also identifies the smallest absolute value in the matrix (`min`).\n\n2. **Handling Negatives**:\n   - If the count of negative elements (`negCount`) is **even**, it\'s possible to flip the signs of the negatives to make all elements positive. In this case, the maximum sum is simply the sum of all absolute values (`sum`).\n   - If `negCount` is **odd**, it\'s impossible to make all elements positive, as flipping an odd number of negatives always leaves one negative. In this case, the algorithm minimizes the impact of the leftover negative by subtracting twice the smallest absolute value from the total sum:  \n     \\[\n     \\text{Adjusted Sum} = \\text{sum} - 2 \\times \\text{min}\n     \\]\n\n3. **Efficiency**:\n   - The code efficiently calculates the desired result in \\(O(n \\times m)\\), where \\(n\\) and \\(m\\) are the dimensions of the matrix.\n\n### Conclusion:\nThe code calculates the **maximum sum achievable by flipping the signs of matrix elements** by leveraging the properties of negative numbers and absolute values. If the number of negatives is even, the total absolute sum is returned. Otherwise, the smallest possible adjustment is made to maximize the result, ensuring correctness and optimality.\n# Code\n```java []\nclass Solution {\n    public long maxMatrixSum(int[][] matrix) {\n    int min = Integer.MAX_VALUE;\n    long sum = 0;\n    int negCount = 0; \n    for(int i=0; i<matrix.length; i++)\n    for(int j=0; j<matrix[0].length; j++)\n    {\n     if(matrix[i][j]<0)\n     negCount++;\n     min = Math.min(min, Math.abs(matrix[i][j]));\n     sum += Math.abs(matrix[i][j]);    \n    }\n    if(negCount%2==0)\n    return sum; \n    return sum - 2*min;\n}\n}\n```
['Array', 'Dynamic Programming', 'Greedy', 'Line Sweep', 'String Matching', 'Matrix', 'Java']