<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,m = len(grid), len(grid[0])
        ans = 0
        if n&1==0 and m&1: # n should be odd if there's odd
            grid = list(zip(*grid))
            grid = [list(grid[i]) for i in range(m)]
            n,m = m,n

        for l in range(n//2):
            r = n-l-1
            for t in range(m//2):
                b = m-t-1
                num1 = sum([grid[l][t], grid[l][b], grid[r][t], grid[r][b]])
                ans += min(num1, 4-num1)
                
        if n&1:
            l = n//2
            if m&1:
                t = m//2
                ans += grid[l][t]
                grid[l][t]=0
                sumxorm = sum([grid[l][j]^grid[l][m-j-1] for j in range(m)])//2
                sumxorn = sum([grid[i][t]^grid[n-i-1][t] for i in range(n)])//2
                sum1 = sum(grid[l])+sum([grid[i][t] for i in range(n)])
                ans += max(min(sum1%4, 4-sum1%4), sumxorm+sumxorn)
            else:
                row = grid[l]
                sumxorm = sum([grid[l][j]^grid[l][m-j-1] for j in range(m)])//2
                sum1 = sum(grid[l])
                # print(sumxorm, sum1)
                ans += max(min(sum1%4, 4-sum1%4), sumxorm)
        return ans 
                    
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