<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://assets.leetcode.com/uploads/2024/05/14/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://assets.leetcode.com/uploads/2024/05/14/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>

class Solution:
    def minimumSum(self, grid: List[List[int]]) -> int:
        def area(r1, r2, c1, c2):
            top, left = sys.maxsize, sys.maxsize
            bot, right = -1, -1
            for row in range(r1, r2):
                for col in range(c1, c2):
                    # print(row, col)
                    if grid[row][col] == 0:
                        continue    
                    top = min(top, row)
                    bot = max(bot, row)
                    left = min(left, col)
                    right = max(right, col)

            return max(right - left + 1, 0) * max(bot - top + 1, 0)

        ans = sys.maxsize
        rows, cols = len(grid), len(grid[0])

        # Row first
        for r in range(1, rows):
            for r2 in range(r + 1, rows):
                a1 = area(0, r, 0, cols)
                a2 = area(r, r2, 0, cols)
                a3 = area(r2, rows, 0, cols)
                ans = min(ans, a1 + a2 + a3)
            
            for c in range(1, cols):
                # print(0, r, 0, cols)
                a1 = area(0, r, 0, cols)
                a2 = area(r, rows, 0, c)
                a3 = area(r, rows, c, cols)
                ans = min(ans, a1 + a2 + a3)

                a1 = area(r, rows, 0, cols)
                a2 = area(0, r, 0, c)
                a3 = area(0, r, c, cols)
                ans = min(ans, a1 + a2 + a3)

        for c in range(1, cols):
            for c2 in range(c + 1, cols):
                a1 = area(0, rows, 0, c)
                a2 = area(0, rows, c, c2)
                a3 = area(0, rows, c2, cols)
                ans = min(ans, a1 + a2 + a3)

            for r in range(1, rows):
                a1 = area(0, rows, 0, c)
                a2 = area(0, r, c, cols)
                a3 = area(r, rows, c, cols)
                ans = min(ans, a1 + a2 + a3)

                a1 = area(0, rows, c, cols)
                a2 = area(0, r, 0, c)
                a3 = area(r, rows, 0, c)
                ans = min(ans, a1 + a2 + a3)

        return ans


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