<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:
        map = defaultdict(int)

        def get_one(i1, j1, i2, j2):
            minx = float('inf')
            maxx = float('-inf')
            miny = float('inf')
            maxy = float('-inf')

            for i in range(i1, i2 + 1):
                for j in range(j1, j2 + 1):
                    if grid[i][j] == 1:
                        minx = min(minx, i)
                        maxx = max(maxx, i)
                        miny = min(miny, j)
                        maxy = max(maxy, j)

            if minx == float('inf'):
                return 0

            return (maxx - minx + 1) * (maxy - miny + 1)

        def get_next(i1, j1, i2, j2, k):
            key = (i1, j1, i2, j2, k)
            if key in map:
                return map[key]

            output = float('inf')

            if k == 1:
                output = get_one(i1, j1, i2, j2)
            elif k == 2:
                for i in range(i1, i2):
                    output = min(output, get_next(i1, j1, i, j2, 1) + get_next(i + 1, j1, i2, j2, 1))
                for j in range(j1, j2):
                    output = min(output, get_next(i1, j1, i2, j, 1) + get_next(i1, j + 1, i2, j2, 1))
            elif k == 3:
                for i in range(i1, i2):
                    output = min(output, get_next(i1, j1, i, j2, 1) + get_next(i + 1, j1, i2, j2, 2))
                    output = min(output, get_next(i1, j1, i, j2, 2) + get_next(i + 1, j1, i2, j2, 1))
                for j in range(j1, j2):
                    output = min(output, get_next(i1, j1, i2, j, 1) + get_next(i1, j + 1, i2, j2, 2))
                    output = min(output, get_next(i1, j1, i2, j, 2) + get_next(i1, j + 1, i2, j2, 1))

            map[key] = output
            return output

        m = len(grid)
        n = len(grid[0])

        ans = get_next(0, 0, m - 1, n - 1, 3)

        return ans
19100