To maximize the sum of the matrix, we should aim to minimize the negative influence. Notice that flipping any pair of adjacent elements will keep the number of negative elements consistent in terms of parity (even or odd). Therefore, the key is to ensure the sum of the negative values is minimized while keeping the matrix's parity even, allowing us to theoretically flip elements to achieve a positive parity. It implies that to maximize the sum, the total count of negative numbers should remain even by flipping the sign of the element with the smallest absolute value.
Time Complexity: O(n^2) because we need to iterate over each element in the matrix.
Space Complexity: O(1) as we use a constant amount of additional space.
1public class MaxMatrixSum {
2 public int maxMatrixSum(int[][] matrix) {
3 int n = matrix.length;
4 int minAbs = Math.abs(matrix[0][0]);
5 int sum = 0;
6 int negCount = 0;
7 for (int i = 0; i < n; i++) {
8 for (int j = 0; j < n; j++) {
9 if (matrix[i][j] < 0) negCount++;
10 sum += Math.abs(matrix[i][j]);
11 minAbs = Math.min(minAbs, Math.abs(matrix[i][j]));
12 }
13 }
14 if (negCount % 2 != 0) {
15 sum -= 2 * minAbs;
16 }
17 return sum;
18 }
19}This Java solution similarly iterates the matrix to compute the overall sum and counts negative elements. By the end, it adjusts the sum by subtracting twice the smallest absolute value if the count of negative elements is odd.
This approach further explores the combination of summing absolute values and preserving parity. Since the act of flipping adjacent pairs maintains the overall parity of negatives, the challenge is to manage the balance of signs dynamically while ensuring operations help convert the majority of negative numbers or maintain equilibrium by pairing onto the minimal interference possible.
Time Complexity: O(n^2); the necessity exists in exploring each point.
Space Complexity: O(1); friendly constant load aside typical iteration encapsulation.
1public class MaximalSum {
2 public int maxMatrixSumOptimized(int[][] matrix) {
3 int n = matrix.length;
4 int minAbsVal = Math.abs(matrix[0][0]);
5 long totalSum = 0;
6 int negativeCount = 0;
7 for (int i = 0; i < n; i++) {
8 for (int j = 0; j < n; j++) {
9 int valueAbs = Math.abs(matrix[i][j]);
10 totalSum += valueAbs;
11 if (matrix[i][j] < 0) negativeCount++;
12 minAbsVal = Math.min(minAbsVal, valueAbs);
13 }
14 }
15 if (negativeCount % 2 != 0) {
16 totalSum -= 2 * minAbsVal;
17 }
18 return (int) totalSum;
19 }
20}The function iterates to compute the overall sum, while keeping a minimal absolute and counting all negative occurrences for parity control. For odd situations, adjust using the least absolute value to ensure equilibrium and maximize the resultant sum.