<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:
    #(x, y) (n - 1 - x, y), (n - 1 - x, m - 1 - y), and (x, m - 1 - y)
    def minFlips(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        ans = 0
        for i in range(n//2):
            for j in range(m//2):
                fourPt = grid[i][j] +  grid[n - i - 1][j] +   grid[i][m - j - 1] +  grid[n - i - 1][m - j - 1]
                ans += min(fourPt, 4-fourPt)
        if n&1 and m&1 and grid[n//2][m//2]:
            ans += 1
        matchOne = 0
        notMatch = 0
        if n&1 :
            i = n // 2
            for j in range(m//2):
                if grid[i][j] == grid[i][m - j - 1]:
                    if grid[i][j] == 1:
                        matchOne += 1
                else:
                    notMatch += 1
        if m&1 :
            j = m // 2
            for i in range(n//2):
                if grid[i][j] == grid[n - i -1][j]:
                    if grid[i][j] == 1:
                        matchOne += 1
                else:
                    notMatch += 1
        if matchOne&1 and notMatch == 0:
            ans += 2
        else:
            ans += notMatch
        return ans
            
                
        
72900