Minimum Number of Flips to Make Binary Grid Palindromic II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3240.Minimum%20Number%20of%20Flips%20to%20Make%20Binary%20Grid%20Palindromic%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3240.Minimum%20Number%20of%20Flips%20to%20Make%20Binary%20Grid%20Palindromic%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3240.Minimum%20Number%20of%20Flips%20to%20Make%20Binary%20Grid%20Palindromic%20II/images/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>
Array; Two Pointers; Matrix
class Solution:
    def minFlips(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ans = 0
        for i in range(m // 2):
            for j in range(n // 2):
                x, y = m - i - 1, n - j - 1
                cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y]
                ans += min(cnt1, 4 - cnt1)
        if m % 2 and n % 2:
            ans += grid[m // 2][n // 2]
        diff = cnt1 = 0
        if m % 2:
            for j in range(n // 2):
                if grid[m // 2][j] == grid[m // 2][n - j - 1]:
                    cnt1 += grid[m // 2][j] * 2
                else:
                    diff += 1
        if n % 2:
            for i in range(m // 2):
                if grid[i][n // 2] == grid[m - i - 1][n // 2]:
                    cnt1 += grid[i][n // 2] * 2
                else:
                    diff += 1
        ans += diff if cnt1 % 4 == 0 or diff else 2
        return ans
