Find the Minimum Area to Cover All Ones II
<p>You are given a 2D <strong>binary</strong> array <code>grid</code>. You need to find 3 <strong>non-overlapping</strong> rectangles having <strong>non-zero</strong> areas with horizontal and vertical sides such that all the 1&#39;s in <code>grid</code> lie inside these rectangles.</p>

<p>Return the <strong>minimum</strong> possible sum of the area of these rectangles.</p>

<p><strong>Note</strong> that the rectangles are allowed to touch.</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 = [[1,0,1],[1,1,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3197.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20II/images/example0rect21.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 280px; height: 198px;" /></p>

<ul>
	<li>The 1&#39;s at <code>(0, 0)</code> and <code>(1, 0)</code> are covered by a rectangle of area 2.</li>
	<li>The 1&#39;s at <code>(0, 2)</code> and <code>(1, 2)</code> are covered by a rectangle of area 2.</li>
	<li>The 1 at <code>(1, 1)</code> is covered by a rectangle of area 1.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1,0,1,0],[0,1,0,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3197.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20II/images/example1rect2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 356px; height: 198px;" /></p>

<ul>
	<li>The 1&#39;s at <code>(0, 0)</code> and <code>(0, 2)</code> are covered by a rectangle of area 3.</li>
	<li>The 1 at <code>(1, 1)</code> is covered by a rectangle of area 1.</li>
	<li>The 1 at <code>(1, 3)</code> is covered by a rectangle of area 1.</li>
</ul>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length, grid[i].length &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either 0 or 1.</li>
	<li>The input is generated such that there are at least three 1&#39;s in <code>grid</code>.</li>
</ul>
Array; Enumeration; Matrix
impl Solution {
    pub fn minimum_sum(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut ans = (m * n) as i32;
        let inf = i32::MAX / 4;

        let f = |i1: usize, j1: usize, i2: usize, j2: usize| -> i32 {
            let mut x1 = inf;
            let mut y1 = inf;
            let mut x2 = -inf;
            let mut y2 = -inf;

            for i in i1..=i2 {
                for j in j1..=j2 {
                    if grid[i][j] == 1 {
                        x1 = x1.min(i as i32);
                        y1 = y1.min(j as i32);
                        x2 = x2.max(i as i32);
                        y2 = y2.max(j as i32);
                    }
                }
            }
            if x1 > x2 || y1 > y2 {
                inf
            } else {
                (x2 - x1 + 1) * (y2 - y1 + 1)
            }
        };

        for i1 in 0..m - 1 {
            for i2 in i1 + 1..m - 1 {
                ans = ans
                    .min(f(0, 0, i1, n - 1) + f(i1 + 1, 0, i2, n - 1) + f(i2 + 1, 0, m - 1, n - 1));
            }
        }

        for j1 in 0..n - 1 {
            for j2 in j1 + 1..n - 1 {
                ans = ans
                    .min(f(0, 0, m - 1, j1) + f(0, j1 + 1, m - 1, j2) + f(0, j2 + 1, m - 1, n - 1));
            }
        }

        for i in 0..m - 1 {
            for j in 0..n - 1 {
                ans = ans.min(f(0, 0, i, j) + f(0, j + 1, i, n - 1) + f(i + 1, 0, m - 1, n - 1));
                ans = ans
                    .min(f(0, 0, i, n - 1) + f(i + 1, 0, m - 1, j) + f(i + 1, j + 1, m - 1, n - 1));
                ans = ans.min(f(0, 0, i, j) + f(i + 1, 0, m - 1, j) + f(0, j + 1, m - 1, n - 1));
                ans = ans
                    .min(f(0, 0, m - 1, j) + f(0, j + 1, i, n - 1) + f(i + 1, j + 1, m - 1, n - 1));
            }
        }

        ans
    }
}
