<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://assets.leetcode.com/uploads/2021/07/23/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>

class Solution:
    def matrixScore(self, grid: List[List[int]]) -> int:
        # 贪心算法 位操作
        # 贪心：首先保证每行第一位为1，然后保证每一列的1尽可能的多
        m, n = len(grid), len(grid[0])
        result = m * (1 << (n-1))   # 左移一位就相当于二进制乘以2

        # 遍历每一列
        for j in range(1, n):
            nOnes = 0
            for i in range(m):
                if grid[i][0] == 1: # 说明没有翻转
                    nOnes += grid[i][j]
                else:   # 翻转了
                    nOnes += 1 - grid[i][j]

            k = max(nOnes, m - nOnes)   # 要使每一列的1个数不少于一半
            result += k * (1 << (n-1-j))

        return result
16000