<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 = len(grid)
        m = len(grid[0])
        ans = 0
        
        if n % 2 == 1 and m % 2 == 1:
            if grid[n // 2][m // 2] == 1:
                ans += 1

        shift = [0] * 3
        if n % 2 == 1:
            a, b = 0, m - 1
            while a < b:
                sm = grid[n // 2][a] + grid[n // 2][b]
                shift[sm] += 1
                a += 1
                b -= 1

        if m % 2 == 1:
            a, b = 0, n - 1
            while a < b:
                sm = grid[a][m // 2] + grid[b][m // 2]
                shift[sm] += 1
                a += 1
                b -= 1

        if shift[2] % 2 == 1:
            if shift[1] > 0:
                ans += shift[1]
            else:
                ans += 2
        else:
            ans += shift[1]

        for i in range(n // 2):
            for j in range(m // 2):
                a = grid[i][j]
                b = grid[n - 1 - i][j]
                c = grid[i][m - 1 - j]
                d = grid[n - 1 - i][m - 1 - j]

                sm = a + b + c + d

                if sm != 0 and sm != 4:
                    ans += min(sm, 4 - sm)

        return ans
        
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