Maximum Matrix Sum
<p>You are given an <code>n x n</code> integer <code>matrix</code>. You can do the following operation <strong>any</strong> number of times:</p>

<ul>
	<li>Choose any two <strong>adjacent</strong> elements of <code>matrix</code> and <strong>multiply</strong> each of them by <code>-1</code>.</li>
</ul>

<p>Two elements are considered <strong>adjacent</strong> if and only if they share a <strong>border</strong>.</p>

<p>Your goal is to <strong>maximize</strong> the summation of the matrix&#39;s elements. Return <em>the <strong>maximum</strong> sum of the matrix&#39;s elements using the operation mentioned above.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1975.Maximum%20Matrix%20Sum/images/pc79-q2ex1.png" style="width: 401px; height: 81px;" />
<pre>
<strong>Input:</strong> matrix = [[1,-1],[-1,1]]
<strong>Output:</strong> 4
<b>Explanation:</b> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1975.Maximum%20Matrix%20Sum/images/pc79-q2ex2.png" style="width: 321px; height: 121px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
<strong>Output:</strong> 16
<b>Explanation:</b> We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 250</code></li>
	<li><code>-10<sup>5</sup> &lt;= matrix[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array; Matrix
class Solution:
    def maxMatrixSum(self, matrix: List[List[int]]) -> int:
        mi = inf
        s = cnt = 0
        for row in matrix:
            for x in row:
                cnt += x < 0
                y = abs(x)
                mi = min(mi, y)
                s += y
        return s if cnt % 2 == 0 else s - mi * 2
