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Author: NamVr

2089-maximum-matrix-sum/README.md

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+<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://assets.leetcode.com/uploads/2021/07/16/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://assets.leetcode.com/uploads/2021/07/16/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>

2089-maximum-matrix-sum/solution.cpp

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+class Solution {
+public:
+    long long maxMatrixSum(vector<vector<int>>& matrix) {
+        // count number of negative signs
+        long long neg_cnt = 0, ans = 0, size = matrix.size(), minn = 1000000;
+
+        // keep track of absolute sum of all values.
+        for (int i = 0; i < size; ++i)
+            for (int j = 0; j < size; ++j) {
+                ans += abs(matrix[i][j]);
+
+                // check if the given element is negative
+                if (matrix[i][j] < 0)
+                    neg_cnt++;
+
+                // keep track of the most minimum value absolute
+                minn = minn < abs(matrix[i][j])? minn: abs(matrix[i][j]);
+            }
+
+        // if negative signs are even, answer is sum of all elements
+        if (neg_cnt % 2 == 0)
+            return ans;
+
+        // otherwise, the most minimum value is subtracted (2x)
+        else 
+            return ans - 2*minn;
+    }
+};