<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>

class Solution:
    def minFlips(self, grid: List[List[int]]) -> int:
    
        n_r, n_c = len(grid), len(grid[0])
        result = 0

        for r in range(n_r//2):
            for c in range(n_c//2):
                cnt = grid[r][c] + grid[n_r-1-r][c] + grid[n_r-1-r][n_c-1-c] + grid[r][n_c-1-c]
                
                if cnt == 2:
                    result += 2
                if cnt == 1 or cnt == 3:
                    result += 1

        extras = 0
        eon = 0
        if n_r % 2:
            for c in range(n_c//2):
                if grid[n_r//2][c] != grid[n_r//2][n_c-1-c]:
                    extras += 1
                elif grid[n_r//2][c]:
                    eon ^= 1
        
        if (n_c % 2):
            for r in range(n_r//2):
                if grid[r][n_c//2] != grid[n_r-1-r][n_c//2]:
                    extras += 1
                elif grid[r][n_c//2]:
                    eon ^= 1
        
        result += extras

        if (eon and not extras):
            result += 2
        if (n_r%2 and n_c%2):
            result += grid[n_r//2][n_c//2]

        return result
        
72800