<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:
        m = len(grid)
        n = len(grid[0])

        ret = 0
        for i in range(m // 2):
            for j in range(n // 2):
                k = grid[i][j] + grid[m - 1 - i][j] + grid[i][n - 1 - j] + grid[m - 1 - i][n - 1 - j]
                ret += min(k, 4 - k)

        if m % 2 == 1:
            if n % 2 == 1:
                if grid[m // 2][n // 2]:
                    ret += 1

                diff = False

                k0 = 0
                for j in range(n // 2):
                    if grid[m // 2][j]:
                        k0 += 1

                    if grid[m // 2][j] != grid[m // 2][n - 1 - j]:
                        diff = True
                        ret += 1

                k1 = 0
                for i in range(m // 2):
                    if grid[i][n // 2]:
                        k1 += 1

                    if grid[i][n // 2] != grid[m - 1 - i][n // 2]:
                        diff = True
                        ret += 1

                if not diff and (k0 + k1) % 2 == 1:
                    ret += 2

            else:
                k = 0
                diff = False
                for j in range(n // 2):
                    if grid[m // 2][j]:
                        k += 1

                    if grid[m // 2][j] != grid[m // 2][n - 1 - j]:
                        diff = True
                        ret += 1

                if not diff and k % 2 == 1:
                    ret += 2

        elif n % 2 == 1:
            k = 0
            diff = False

            for i in range(m // 2):
                if grid[i][n // 2]:
                    k += 1

                if grid[i][n // 2] != grid[m - 1 - i][n // 2]:
                    diff = True
                    ret += 1

            if not diff and k % 2 == 1:
                ret += 2

        return ret
72100