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
function minFlips(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;

    for (let i = 0; i < Math.floor(m / 2); i++) {
        for (let j = 0; j < Math.floor(n / 2); j++) {
            const x = m - i - 1;
            const y = n - j - 1;
            const cnt1 = grid[i][j] + grid[x][j] + grid[i][y] + grid[x][y];
            ans += Math.min(cnt1, 4 - cnt1);
        }
    }

    if (m % 2 === 1 && n % 2 === 1) {
        ans += grid[Math.floor(m / 2)][Math.floor(n / 2)];
    }

    let diff = 0,
        cnt1 = 0;

    if (m % 2 === 1) {
        for (let j = 0; j < Math.floor(n / 2); j++) {
            if (grid[Math.floor(m / 2)][j] === grid[Math.floor(m / 2)][n - j - 1]) {
                cnt1 += grid[Math.floor(m / 2)][j] * 2;
            } else {
                diff += 1;
            }
        }
    }

    if (n % 2 === 1) {
        for (let i = 0; i < Math.floor(m / 2); i++) {
            if (grid[i][Math.floor(n / 2)] === grid[m - i - 1][Math.floor(n / 2)]) {
                cnt1 += grid[i][Math.floor(n / 2)] * 2;
            } else {
                diff += 1;
            }
        }
    }

    ans += cnt1 % 4 === 0 || diff > 0 ? diff : 2;
    return ans;
}
