Leetcode 1975: Maximum Matrix Sum
You are given a square matrix of integers. You can perform the following operation multiple times: choose any two adjacent elements in the matrix and multiply both of them by -1. Your goal is to maximize the sum of the matrix elements after performing the operation any number of times.
Problem
Approach
Steps
Complexity
Input: The input consists of a square matrix of integers `matrix` with `n` rows and `n` columns (2 <= n <= 250).
Example: matrix = [[1, -2], [-3, 4]]
Constraints:
• 2 <= n <= 250
• -10^5 <= matrix[i][j] <= 10^5
Output: The output is a single integer, representing the maximum sum of the matrix after performing the operation any number of times.
Example: Output: 16
Constraints:
• The result should be an integer.
Goal: Maximize the sum of the matrix elements by applying the given operation.
Steps:
• Step 1: Calculate the sum of the absolute values of all elements in the matrix.
• Step 2: Count the number of negative elements in the matrix.
• Step 3: Identify the minimum absolute value element in the matrix.
• Step 4: If the number of negative elements is odd, subtract twice the minimum absolute value from the total sum to maximize the sum.
• Step 5: Return the final sum.
Goal: The solution must efficiently handle matrices of size up to 250x250.
Steps:
• The matrix will always be square with n rows and n columns.
• The elements in the matrix are integers and can be negative.
Assumptions:
• The matrix is always square, i.e., the number of rows is equal to the number of columns.
• The matrix contains integers within the given bounds.
• Input: Input: matrix = [[1, -2], [-3, 4]]
• Explanation: In this case, the sum of absolute values is |1| + |-2| + |-3| + |4| = 10. The number of negative elements is 2, which is even. Therefore, we do not need to adjust the sum. The maximum sum is 10.
• Input: Input: matrix = [[-1, 2], [-3, -4]]
• Explanation: In this case, the sum of absolute values is |-1| + |2| + |-3| + |-4| = 10. The number of negative elements is 3, which is odd. We subtract twice the minimum absolute value (which is 1) from the sum, resulting in 10 - 2 = 8. The maximum sum is 8.
Approach: To solve this problem, we need to compute the sum of absolute values and then adjust the sum if necessary based on the number of negative elements.
Observations:
• The matrix's elements can be negative, so we should consider the absolute values to maximize the sum.
• If the number of negative elements is odd, we need to adjust the total sum by subtracting twice the smallest absolute value.
• The solution should compute the sum of absolute values first, then adjust the sum based on the number of negative elements.
Steps:
• Step 1: Initialize a variable to hold the total sum of the absolute values of all elements in the matrix.
• Step 2: Count the number of negative elements in the matrix.
• Step 3: Find the smallest absolute value in the matrix.
• Step 4: If the number of negative elements is odd, subtract twice the smallest absolute value from the total sum.
• Step 5: Return the final sum.
Empty Inputs:
• The matrix will always have at least 2 rows and 2 columns.
Large Inputs:
• The solution must efficiently handle matrices as large as 250x250.
Special Values:
• If all elements in the matrix are positive, no adjustments are needed.
• If all elements are negative and the number of negative elements is odd, the sum will be adjusted.
Constraints:
• Ensure the solution handles the matrix size and the range of integer values correctly.
long long maxMatrixSum(vector<vector<int>>& mat) {
int n = mat.size();
long long sum = 0;
bool isodd = n % 2;
int cnt = 0;
int mn = abs(mat[0][0]);
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++) {
if(mat[i][j] < 0) cnt++;
mn = min(mn, abs(mat[i][j]));
sum += abs(mat[i][j]);
}
if(cnt%2 == 1)
sum -= (2 *mn);
return sum;
}
1 : Function Definition
long long maxMatrixSum(vector<vector<int>>& mat) {
This is the function definition for 'maxMatrixSum', which calculates the maximum sum of absolute values in a matrix while handling negative elements.
2 : Matrix Size Calculation
int n = mat.size();
This line calculates the size of the matrix 'mat' and stores it in the variable 'n'. The matrix is assumed to be square.
3 : Variable Initialization
long long sum = 0;
The 'sum' variable is initialized to zero, and it will hold the cumulative sum of absolute values of matrix elements.
4 : Odd Check
bool isodd = n % 2;
The variable 'isodd' checks whether the size of the matrix 'n' is odd or even. This variable is used later for adjustments based on the number of negative values.
5 : Negative Count Initialization
int cnt = 0;
The variable 'cnt' is initialized to count the number of negative elements in the matrix.
6 : Minimum Value Initialization
int mn = abs(mat[0][0]);
The 'mn' variable is initialized to the absolute value of the first element of the matrix. It will keep track of the smallest absolute value in the matrix.
7 : Outer Loop
for(int i = 0; i < n; i++)
This is the outer loop, which iterates over each row of the matrix.
8 : Inner Loop
for(int j = 0; j < n; j++) {
This is the inner loop, which iterates over each element in the current row of the matrix.
9 : Negative Count Update
if(mat[i][j] < 0) cnt++;
If the current element in the matrix is negative, increment the 'cnt' variable to keep track of the number of negative elements.
10 : Minimum Value Update
mn = min(mn, abs(mat[i][j]));
Update the 'mn' variable to store the minimum absolute value encountered in the matrix so far.
11 : Sum Update
sum += abs(mat[i][j]);
Add the absolute value of the current matrix element to the 'sum'.
12 : Odd Negative Count Adjustment
if(cnt%2 == 1)
If the number of negative elements is odd, an adjustment is needed to maximize the sum.
13 : Adjustment to Sum
sum -= (2 *mn);
If the number of negative elements is odd, subtract twice the minimum absolute value from the sum to make the total sum even.
14 : Return Statement
return sum;
Return the computed sum, which is the maximum possible sum of absolute values in the matrix.
Best Case: O(n^2), where `n` is the number of rows (or columns) in the matrix.
Average Case: O(n^2)
Worst Case: O(n^2)
Description: We need to iterate through all elements of the matrix to calculate the sum and count the negatives, which takes O(n^2) time.
Best Case: O(1)
Worst Case: O(1)
Description: We only need a constant amount of extra space to store the sum, count, and minimum value, so the space complexity is O(1).
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