<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:            
        R, C = len(grid), len(grid[0])
        ones, changes = 0, 0

        for r in range((R)//2):
            for c in range((C)//2):
                v1 = grid[r][c]
                v2 = grid[r][C-c-1]
                v3 = grid[R-r-1][c]
                v4 = grid[R-r-1][C-c-1]
                
                if v1 + v2 + v3 + v4 > 2:
                    changes += 4 - (v1 + v2 + v3 + v4)
                else:
                    changes += (v1 + v2 + v3 + v4)

        pairs, single = 0, 0
        if R % 2:
            for c in range(C//2):
                if grid[R//2][c] != grid[R//2][C-c-1]:
                    single += 1
                elif grid[R//2][c]:
                    pairs += 2

        if C % 2:
            for r in range(R//2):
                if grid[r][C//2] != grid[R-r-1][C//2]:
                    single += 1
                elif grid[r][C//2] == 1:
                    pairs += 2

        if (R % 2) and (C % 2):
            if grid[R//2][C//2]:
                changes += 1

        if pairs % 4 == 0 or single > 0:
            changes += single
        else:
            changes += 2
        return changes
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