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
func minFlips(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	ans := 0

	for i := 0; i < m/2; i++ {
		for j := 0; j < n/2; j++ {
			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 == 1 && n%2 == 1 {
		ans += grid[m/2][n/2]
	}

	diff, cnt1 := 0, 0

	if m%2 == 1 {
		for j := 0; j < n/2; j++ {
			if grid[m/2][j] == grid[m/2][n-j-1] {
				cnt1 += grid[m/2][j] * 2
			} else {
				diff += 1
			}
		}
	}

	if n%2 == 1 {
		for i := 0; i < m/2; i++ {
			if grid[i][n/2] == grid[m-i-1][n/2] {
				cnt1 += grid[i][n/2] * 2
			} else {
				diff += 1
			}
		}
	}

	if cnt1%4 == 0 || diff > 0 {
		ans += diff
	} else {
		ans += 2
	}

	return ans
}
