<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, n = len(grid), len(grid[0])
        visited = set()
        pairs = []

        def get_cycle(i, j):
            ones, zeros = 0, 0
            if grid[i][j] == 1:
                ones += 1
            else:
                zeros += 1
            q = deque()
            q.append((i, j))
            visited.add((i, j))
            while q:
                cx, cy = q.popleft()
                ox, oy = m - 1 - cx, n - 1 - cy

                if (cx, oy) not in visited:
                    visited.add((cx, oy))
                    q.append((cx, oy))
                    if grid[cx][oy] == 1:
                        ones += 1
                    else:
                        zeros += 1
                
                if (ox, cy) not in visited:
                    visited.add((ox, cy))
                    q.append((ox, cy))
                    if grid[ox][cy] == 1:
                        ones += 1
                    else:
                        zeros += 1

            return ones, zeros
        
        def calculate_flips():
            for i in range((m + 1) // 2):
                for j in range((n + 1) // 2):
                    if (i, j) not in visited:
                        ones, zeros = get_cycle(i, j)
                        pairs.append((ones, zeros))

        calculate_flips()
        s_flips = 0
        zc, oc, eq = 0, 0, 0
        for ii, jj in pairs:
            if (ii + jj) == 4:
                s_flips += min(ii, jj)
            elif (ii + jj) == 2:
                if ii == jj:
                    eq+=1
                elif ii > jj:
                    oc += 1
                else:
                    zc += 1
            else:
                if ii == 1:
                    s_flips += 1
        if oc%2 == 0:
            return s_flips + eq
        rc = oc%2
        if eq >= rc:
            return s_flips + eq
        rc -= eq
        return s_flips + (rc*2)
81000