<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>

from typing import List

class Solution:
    def minimumArea(self, grid: List[List[int]], st_i: int, en_i: int, st_j: int, en_j: int) -> int:
        top, bottom, left, right = -1, -1, -1, -1
        
        for i in range(st_i, en_i + 1):
            for j in range(st_j, en_j + 1):
                if grid[i][j] == 1:
                    if top == -1:
                        top = i
                    bottom = i
        
        for j in range(st_j, en_j + 1):
            for i in range(st_i, en_i + 1):
                if grid[i][j] == 1:
                    if left == -1:
                        left = j
                    right = j

        # If no 1 was found, area should be 0
        if top == -1 or left == -1:
            return 0

        return (right - left + 1) * (bottom - top + 1)
    
    def minimumSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ans = int(1e9)
        
        # Pattern 1
        for i in range(m):
            one = self.minimumArea(grid, 0, i, 0, n - 1)
            for j in range(n):
                two = self.minimumArea(grid, i + 1, m - 1, 0, j)
                three = self.minimumArea(grid, i + 1, m - 1, j + 1, n - 1)
                ans = min(ans, one + two + three)
        
        # Pattern 2
        for j in range(n):
            one = self.minimumArea(grid, 0, m - 1, 0, j)
            for i in range(m):
                two = self.minimumArea(grid, 0, i, j + 1, n - 1)
                three = self.minimumArea(grid, i + 1, m - 1, j + 1, n - 1)
                ans = min(ans, one + two + three)
        
        # Pattern 3
        for j in range(n - 1, -1, -1):
            one = self.minimumArea(grid, 0, m - 1, j + 1, n - 1)
            for i in range(m):
                two = self.minimumArea(grid, 0, i, 0, j)
                three = self.minimumArea(grid, i + 1, m - 1, 0, j)
                ans = min(ans, one + two + three)
        
        # Pattern 4
        for i in range(m - 1, -1, -1):
            one = self.minimumArea(grid, i + 1, m - 1, 0, n - 1)
            for j in range(n):
                two = self.minimumArea(grid, 0, i, 0, j)
                three = self.minimumArea(grid, 0, i, j + 1, n - 1)
                ans = min(ans, one + two + three)
        
        # Pattern 5
        for i in range(m):
            for j in range(i + 1, m):
                one = self.minimumArea(grid, 0, i, 0, n - 1)
                two = self.minimumArea(grid, i + 1, j, 0, n - 1)
                three = self.minimumArea(grid, j + 1, m - 1, 0, n - 1)
                ans = min(ans, one + two + three)
        
        # Pattern 6
        for i in range(n):
            for j in range(i + 1, n):
                one = self.minimumArea(grid, 0, m - 1, 0, i)
                two = self.minimumArea(grid, 0, m - 1, i + 1, j)
                three = self.minimumArea(grid, 0, m - 1, j + 1, n - 1)
                ans = min(ans, one + two + three)
        
        return ans

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