Introduction

In this blog post, we will explore how to solve the LeetCode problem "1975" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution:

• Key Idea: The core idea is to realize that we can effectively move a negative sign around the matrix. The only thing that matters is the total number of negative numbers and the minimum absolute value of an element. If the count of negative numbers is even, we can make all elements positive. If it's odd, we'll have one negative element remaining, and it's optimal to make the element with the smallest absolute value negative.

• Count Negatives and Find Minimum Absolute Value: Iterate through the matrix, count the number of negative elements, and keep track of the minimum absolute value of all elements.

• Calculate Sum: If the negative count is even, the maximum sum is simply the sum of the absolute values of all elements. If the count is odd, we subtract twice the minimum absolute value from the sum of absolute values (since we're essentially flipping the sign of the smallest absolute value element).

• Complexity:

  • Runtime: O(n^2)
  • Storage: O(1)

Code

    def maxMatrixSum(matrix):
    n = len(matrix)
    negative_count = 0
    min_abs = float('inf')
    total_sum = 0

    for i in range(n):
        for j in range(n):
            if matrix[i][j] < 0:
                negative_count += 1
            min_abs = min(min_abs, abs(matrix[i][j]))
            total_sum += abs(matrix[i][j])

    if negative_count % 2 == 0:
        return total_sum
    else:
        return total_sum - 2 * min_abs