Daily LeetCode - by Vítor Belim

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

This final version fixes the earlier mistake by tracking the right value to sacrifice. In a single O(m × n) pass it adds every cell's magnitude to total (treating each value as if it could be made positive), counts non-positive entries in negativeCounter, and — crucially — records the smallest absolute value across the whole matrix in smallestNumber, checking Math.abs(matrix[r][c]) for every cell rather than only the negative ones. Because the allowed operation flips two adjacent values at once, an even number of negatives can always be cancelled completely; if the bitwise test negativeCounter & 1 reveals an odd count, exactly one value must end up negative, and the cheapest choice is the one with the smallest magnitude, so smallestNumber * 2 is subtracted (its contribution swings from positive to negative). This greedy approach solves the problem in one pass with constant extra space.

Language: javascript(2026-01-05 10:12)DONE

CPU Performance100.00%

Memory Performance94.12%

/**
 * @param {number[][]} matrix
 * @return {number}
 */
var maxMatrixSum = function(matrix) {
    const m = matrix.length;
    const n = matrix[0].length;

    let total = 0;
    let negativeCounter = 0;
    let smallestNumber = Infinity;

    for (let r = 0; r < m; r++) {
        for (let c = 0; c < n; c++) {
            if (matrix[r][c] > 0) {
                total += matrix[r][c];
            } else {
                total += -matrix[r][c];
                negativeCounter++;
            }

            smallestNumber = Math.min(smallestNumber, Math.abs(matrix[r][c]));
        }
    }

    if (negativeCounter & 1) {
        total -= smallestNumber * 2;
    }

    return total;
};

Description

Solutions

AI Explanation