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

Python3JavaC++GoJavaScript

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

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;
    }
}

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;
    }
};

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
}

/**
 * @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;
};

1975. Maximum Matrix Sum

Description

Solutions

Solution 1

Comments