<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:
        #Connect 4 squares
        #Set to min(cost(0), cost(1))
        
        #2 options:
        #"corner" cells (treat normally)
        #center row/col
        #Either %4==2 or %4==0
        #Test min for 2 for both or min for 4 for both, take min
        
        m, n = len(grid), len(grid[0])
        res = 0
        #Corners
        for i in range(m//2):
            for j in range(n//2):
                x = grid[i][j]+grid[m-1-i][j]+grid[m-1-i][n-1-j]+grid[i][n-1-j]
                res+=min(x, 4-x)
        #Centers
        row, col = None, None
        marked, offs = 0, 0
        if m&1:
            row = grid[m//2]
            x = sum([row[i] != row[n-1-i] for i in range(n//2)])
            marked|=bool(x)
            offs+=sum(row[i]+row[n-1-i] for i in range(n//2))
            res+=x
        if n&1:
            col = [grid[i][n//2] for i in range(m)]
            x = sum([col[i] != col[m-1-i] for i in range(m//2)])
            marked|=bool(x)
            offs+=sum(col[i]+col[m-1-i] for i in range(m//2))
            res+=x
        
        if m&1 or n&1:
            #Make sure %4==0
            # print(res, offs, marked)
            res+=(offs%4 and not marked)*2
            #Make sure center is deleted
            if m&1 and n&1: res+=grid[m//2][n//2]
        return res
71500