Score After Flipping Matrix
<p>You are given an <code>m x n</code> binary matrix <code>grid</code>.</p>

<p>A <strong>move</strong> consists of choosing any row or column and toggling each value in that row or column (i.e., changing all <code>0</code>&#39;s to <code>1</code>&#39;s, and all <code>1</code>&#39;s to <code>0</code>&#39;s).</p>

<p>Every row of the matrix is interpreted as a binary number, and the <strong>score</strong> of the matrix is the sum of these numbers.</p>

<p>Return <em>the highest possible <strong>score</strong> after making any number of <strong>moves</strong> (including zero moves)</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/0800-0899/0861.Score%20After%20Flipping%20Matrix/images/lc-toogle1.jpg" style="width: 500px; height: 299px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
<strong>Output:</strong> 39
<strong>Explanation:</strong> 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 20</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Greedy; Bit Manipulation; Array; Matrix
public class Solution {
    public int MatrixScore(int[][] grid) {
        int m = grid.Length, n = grid[0].Length;
        for (int i = 0; i < m; ++i) {
            if (grid[i][0] == 0) {
                for (int j = 0; j < n; ++j) {
                    grid[i][j] ^= 1;
                }
            }
        }
        int ans = 0;
        for (int j = 0; j < n; ++j) {
            int cnt = 0;
            for (int i = 0; i < m; ++i) {
                if (grid[i][j] == 1) {
                    ++cnt;
                }
            }
            ans += Math.Max(cnt, m - cnt) * (1 << (n - j - 1));
        }
        return ans;
    }
}
