1975. Maximum Matrix Sum

Description

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

Solutions

Solution 1

Python Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

class Solution:
    def maxMatrixSum(self, matrix: List[List[int]]) -> int:
        s = cnt = 0
        mi = inf
        for row in matrix:
            for v in row:
                s += abs(v)
                mi = min(mi, abs(v))
                if v < 0:
                    cnt += 1
        if cnt % 2 == 0 or mi == 0:
            return s
        return s - mi * 2

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

class Solution {
    public long maxMatrixSum(int[][] matrix) {
        long s = 0;
        int cnt = 0;
        int mi = Integer.MAX_VALUE;
        for (var row : matrix) {
            for (var v : row) {
                s += Math.abs(v);
                mi = Math.min(mi, Math.abs(v));
                if (v < 0) {
                    ++cnt;
                }
            }
        }
        if (cnt % 2 == 0 || mi == 0) {
            return s;
        }
        return s - mi * 2;
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
public:
    long long maxMatrixSum(vector<vector<int>>& matrix) {
        long long s = 0;
        int cnt = 0, mi = INT_MAX;
        for (auto& row : matrix) {
            for (int& v : row) {
                s += abs(v);
                mi = min(mi, abs(v));
                cnt += v < 0;
            }
        }
        if (cnt % 2 == 0 || mi == 0) return s;
        return s - mi * 2;
    }
};

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

func maxMatrixSum(matrix [][]int) int64 {
	s := 0
	cnt, mi := 0, math.MaxInt32
	for _, row := range matrix {
		for _, v := range row {
			s += abs(v)
			mi = min(mi, abs(v))
			if v < 0 {
				cnt++
			}
		}
	}
	if cnt%2 == 1 {
		s -= mi * 2
	}
	return int64(s)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

JavaScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

/**
 * @param {number[][]} matrix
 * @return {number}
 */
var maxMatrixSum = function (matrix) {
    let cnt = 0;
    let s = 0;
    let mi = Infinity;
    for (const row of matrix) {
        for (const v of row) {
            s += Math.abs(v);
            mi = Math.min(mi, Math.abs(v));
            cnt += v < 0;
        }
    }
    if (cnt % 2 == 0) {
        return s;
    }
    return s - mi * 2;
};