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

<p>A row or column is considered <strong>palindromic</strong> if its values read the same forward and backward.</p>

<p>You can <strong>flip</strong> any number of cells in <code>grid</code> from <code>0</code> to <code>1</code>, or from <code>1</code> to <code>0</code>.</p>

<p>Return the <strong>minimum</strong> number of cells that need to be flipped to make <strong>all</strong> rows and columns <strong>palindromic</strong>, and the total number of <code>1</code>&#39;s in <code>grid</code> <strong>divisible</strong> by <code>4</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1,0,0],[0,1,0],[0,0,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p><img src="https://assets.leetcode.com/uploads/2024/08/01/image.png" style="width: 400px; height: 105px;" /></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[0,1],[0,1],[0,0]]</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/07/08/screenshot-from-2024-07-09-01-37-48.png" style="width: 300px; height: 104px;" /></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1],[1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/01/screenshot-from-2024-08-01-23-05-26.png" style="width: 200px; height: 70px;" /></p>
</div>

<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;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 1</code></li>
</ul>

from typing import List

class Solution:
    def minFlips(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        res = single = double = 0

        def count_ones(i, j):
            return grid[i][j] + grid[i][n-1-j] + grid[m-1-i][j] + grid[m-1-i][n-1-j]

        def count_single_double(i, j):
            ones = grid[i][j] + grid[m-1-i][j]
            return (ones == 1), (ones == 2)

        for i in range(m // 2):
            for j in range(n // 2):
                ones = count_ones(i, j)
                res += min(ones, 4 - ones)

            if n % 2 == 1:
                ones = grid[i][n // 2] + grid[m - 1 - i][n // 2]
                single += (ones == 1)
                double += (ones == 2)

        if m % 2 == 1:
            for j in range(n // 2):
                ones = grid[m // 2][j] + grid[m // 2][n - 1 - j]
                single += (ones == 1)
                double += (ones == 2)
            if n % 2 == 1:
                res += grid[m // 2][n // 2]

        if double % 2 == 0 or single > 0:
            return res + single
        return res + 2

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