<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])
        n2, m2 = math.ceil(n/2), math.ceil(m/2)
        dp = [[[-1 for _ in range(4)] for _ in range(m2)] for _ in range(n2)]
        dp[-1].append([0, float('inf'), float('inf'), float('inf')])
        initSum = sum(sum(row) for row in grid) % 4
        for i in range(n2-1, -1, -1):
            for j in range(m2-1, -1, -1):
                nxtArray = dp[i][j+1] if j != len(dp[i])-1 else dp[i+1][0]
                cons = [(i, j), (i, m - j - 1), (n - i - 1, j), (n - i - 1, m - j - 1)]
                seen = set()
                ones = totChanging = 0
                for ri, rj in cons:
                    t = (ri, rj)
                    if t not in seen:
                        seen.add(t)
                        totChanging += 1
                        ones += grid[ri][rj]

                for s in range(4):
                    # currently looking at (i, j) with modular sum s
                    changeToOnes = totChanging - ones
                    changeToZeroes = ones
                    dp[i][j][s] = min(changeToOnes + nxtArray[(s + changeToOnes) % 4], \
                                      changeToZeroes + nxtArray[(s - changeToZeroes + 4) % 4])
        return dp[0][0][initSum]
71900