<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:
        import math
        
        m, n = len(grid), len(grid[0])
        ans = 0

        
        for i in range(m // 2):
            for j in range(n // 2):
                cnt = 0
                cnt += grid[i][j]
                cnt += grid[m - 1 - i][j]
                cnt += grid[i][n - 1 - j]
                cnt += grid[m - 1 - i][n - 1 - j]
                ans += min(cnt, 4 - cnt)
        
    
        diff = 0
        p0 = 0
        p1 = 0
        
        if m % 2 == 1:
            mid_row = m // 2
            for j in range(n // 2):
                if grid[mid_row][j] != grid[mid_row][n - j - 1]:
                    diff += 1
                else:
                    if grid[mid_row][j] == 0:
                        p0 += 1
                    else:
                        p1 += 1
        
        if n % 2 == 1:
            mid_col = n // 2
            for i in range(m // 2):
                if grid[i][mid_col] != grid[m - i - 1][mid_col]:
                    diff += 1
                else:
                    if grid[i][mid_col] == 0:
                        p0 += 1
                    else:
                        p1 += 1
        
        if n % 2 == 1 and m % 2 == 1:
            if grid[m // 2][n // 2] == 1:
                ans += 1
        
        ans1 = math.inf
        if diff % 2 == p1 % 2:
            ans1 = diff
        else:
            if diff % 2 == 0:
                if diff == 0:
                    ans1 = 2
                else:
                    ans1 = diff
            else:
                ans1 = diff
        
        return ans + ans1
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