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Maximize matrix sum by repeatedly multiplying pairs of adjacent elements with -1

 1 year ago
source link: https://www.geeksforgeeks.org/maximize-matrix-sum-by-repeatedly-multiplying-pairs-of-adjacent-elements-with-1/
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Maximize matrix sum by repeatedly multiplying pairs of adjacent elements with -1

Given a matrix A[][] of dimensions M × N, the task is to find the maximum sum possible from a matrix by repeatedly selecting two adjacent matrix elements and multiplying both of their values with -1.

Examples:

Input: A[ ][ ] = {{4, -8, 6}, {3, 7, 2}}
Output: 26
Explanation:
Multiply mat[0][1] and mat[0][2] by -1. The matrix modifies to A[][] = {{4, 8, -6}, {3, 7, 2}}.
Multiply mat[0][2] and mat[1][2] by -1. The matrix modifies to A[][] = {{4, 8, 6}, {3, 7, -2}}.
Therefore, sum of the matrix = 26, which is the maximum sum possible.

Input: A[ ][ ] = {{2, 9, -5}, {6, 1, 3}, {-7, 4, 8}}
Output: 45
Explanation:
Multiply mat[0][2] and mat[1][2] by -1. The matrix modifies to A[][] = {{2, 9, 5}, {6, 1, -3}, {-7, 4, 8}}.
Multiply mat[1][1] and mat[1][2] by -1. The matrix modifies to A[][] = {{2, 9, 5}, {6, -1, 3}, {-7, 4, 8}}.
Multiply mat[1][1] and mat[2][1] by -1. The matrix modifies to A[][] = {{2, 9, 5}, {6, 1, 3}, {-7, -4, 8}}.
Multiply mat[2][0] and mat[2][1] by -1. The matrix modifies to A[][] = {{2, 9, 5}, {6, 1, 3}, {7, 4, 8}}.
Therefore, sum of the matrix = 45, which is the maximum sum possible.

Approach: To maximize the sum of the given matrix, perform the given operations such that the smallest element in a row or column is negative (if it is not possible to make all row and column elements positive). Follow the steps below to maximize the sum:

  • Let the number of cells having negative values be x. If x is 0 i.e., there are no negative values, then the sum of the matrix is already maximum possible.
  • Otherwise, take any negative value from the matrix and perform the given operations on that element and any of its adjacent cells. This results in the chosen element becoming positive.
  • Now if the value of the chosen adjacent element is negative, then it will become positive or vice versa. This way, in each turn two negative values can be made positive. Now following two cases arise:
    • If x is even: In this case, after x/2 turns, all the negative values can be made positive. Therefore, the maximum sum possible is equal to the sum of the absolute values of all cells.
    • If x is odd: In this case, there will be one negative value left after performing operations in the manner described above. Now, to maximize the sum, the value getting subtracted or having a negative sign needs to be minimized. To do this, move the negative sign to the cell having minimum absolute value, say minVal. Therefore, the maximum sum possible is sum of the absolute values of all cells – 2*minVal.

Below is the implementation of the above approach:

  • Python3
  • Javascript
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate maximum sum
// possible of a matrix by multiplying
// pairs of adjacent elements with -1
// any number of times (possibly zero)
void getMaxSum(vector<vector<int> > A,
int M, int N)
{
// Store the maximum sum
// of matrix possible
int sum = 0;
// Stores the count of
// negative values in the matrix
int negative = 0;
// Store minimum absolute
// value present in the matrix
int minVal = INT_MAX;
// Traverse the matrix row-wise
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// Update sum
sum += abs(A[i][j]);
// If current element is negative,
// increment the negative count
if (A[i][j] < 0) {
negative++;
}
// If current value is less than
// the overall minimum in A[],
// update the overall minimum
minVal = min(minVal, abs(A[i][j]));
}
}
// If there are odd number of negative
// values, then the answer will be
// sum of the absolute values of
// all elements - 2* minVal
if (negative % 2) {
sum -= 2 * minVal;
}
// Print maximum sum
cout << sum;
}
// Driver Code
int main()
{
// Given matrix
vector<vector<int> > A = {
{ 4, -8, 6 },
{ 3, 7, 2 }
};
// Dimensions of matrix
int M = A.size();
int N = A[0].size();
getMaxSum(A, M, N);
return 0;
}
Output: 
26

Time Complexity: O(M*N)
Auxiliary Space: O(1)


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