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
class Solution {
public:
    int minimumSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = m * n;
        int inf = INT_MAX / 4;
        auto f = [&](int i1, int j1, int i2, int j2) {
            int x1 = inf, y1 = inf;
            int x2 = -inf, y2 = -inf;
            for (int i = i1; i <= i2; i++) {
                for (int j = j1; j <= j2; j++) {
                    if (grid[i][j] == 1) {
                        x1 = min(x1, i);
                        y1 = min(y1, j);
                        x2 = max(x2, i);
                        y2 = max(y2, j);
                    }
                }
            }
            return x1 > x2 || y1 > y2 ? inf : (x2 - x1 + 1) * (y2 - y1 + 1);
        };

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

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

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

        return ans;
    }
};