1975. Maximum Matrix Sum
Difficulty: Medium
Topics: Array, Greedy, Matrix
You are given an n x n integer matrix. You can do the following operation any number of times:
- Choose any two adjacent elements of matrix and multiply each of them by -1.
Two elements are considered adjacent if and only if they share a border.
Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above.
Example 1:
- Input: matrix = [[1,-1],[-1,1]]
- Output: 4
-
Explanation: We can follow the following steps to reach sum equals 4:
- Multiply the 2 elements in the first row by -1.
- Multiply the 2 elements in the first column by -1.
Example 2:
- Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
- Output: 16
-
Explanation: We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.
Constraints:
- n == matrix.length == matrix[i].length
- 2 <= n <= 250
- -105 <= matrix[i][j] <= 105
Hint:
- Try to use the operation so that each row has only one negative number.
- If you have only one negative element you cannot convert it to positive.
Solution:
To maximize the matrix's sum using the operation, we need to minimize the absolute value of the negative contributions to the sum. Here's the plan:
- Count Negative Numbers: Track how many negative numbers are present in the matrix.
- Find Minimum Absolute Value: Determine the smallest absolute value in the matrix, which will help if the number of negatives is odd.
- Adjust for Odd Negatives: If the count of negative numbers is odd, the smallest absolute value cannot be flipped to positive, and this limits the maximum possible sum.
Let's implement this solution in PHP: 1975. Maximum Matrix Sum
Explanation:
- Summation of Absolute Values: Compute the sum of absolute values of all elements since the optimal configuration flips as many negative numbers to positive as possible.
- Track Smallest Absolute Value: Use this to adjust the sum when the count of negative numbers is odd.
- Handle Odd Negatives: Subtract twice the smallest absolute value from the sum to account for the unavoidable negative element when the negatives cannot be fully neutralized.
Complexity
- Time Complexity: O(n^2), as the matrix is traversed once.
- Space Complexity: O(1), since no additional space is used apart from variables.
This solution works efficiently within the given constraints.
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