<p>You are given an <code>m x n</code> integer matrix <code>grid</code>​​​.</p>

<p>A <strong>rhombus sum</strong> is the sum of the elements that form <strong>the</strong> <strong>border</strong> of a regular rhombus shape in <code>grid</code>​​​. The rhombus must have the shape of a square rotated 45 degrees with each of the corners centered in a grid cell. Below is an image of four valid rhombus shapes with the corresponding colored cells that should be included in each <strong>rhombus sum</strong>:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/23/pc73-q4-desc-2.png" style="width: 385px; height: 385px;" />
<p>Note that the rhombus can have an area of 0, which is depicted by the purple rhombus in the bottom right corner.</p>

<p>Return <em>the biggest three <strong>distinct rhombus sums</strong> in the </em><code>grid</code><em> in <strong>descending order</strong></em><em>. If there are less than three distinct values, return all of them</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/23/pc73-q4-ex1.png" style="width: 360px; height: 361px;" />
<pre>
<strong>Input:</strong> grid = [[3,4,5,1,3],[3,3,4,2,3],[20,30,200,40,10],[1,5,5,4,1],[4,3,2,2,5]]
<strong>Output:</strong> [228,216,211]
<strong>Explanation:</strong> The rhombus shapes for the three biggest distinct rhombus sums are depicted above.
- Blue: 20 + 3 + 200 + 5 = 228
- Red: 200 + 2 + 10 + 4 = 216
- Green: 5 + 200 + 4 + 2 = 211
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/23/pc73-q4-ex2.png" style="width: 217px; height: 217px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [20,9,8]
<strong>Explanation:</strong> The rhombus shapes for the three biggest distinct rhombus sums are depicted above.
- Blue: 4 + 2 + 6 + 8 = 20
- Red: 9 (area 0 rhombus in the bottom right corner)
- Green: 8 (area 0 rhombus in the bottom middle)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[7,7,7]]
<strong>Output:</strong> [7]
<strong>Explanation:</strong> All three possible rhombus sums are the same, so return [7].
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def getBiggestThree(self, M: List[List[int]]) -> List[int]:
        m = len(M)
        n = len(M[0])

        # prfix sum:  additional 1 top row, 1 left col
        dagl_pre_sum = [[0] * (1 + n) for _ in range(1 + m)]
        skew_pre_sum = [[0] * (1 + n) for _ in range(1 + m)]

        for i in range(m):
            for j in range(n):
                dagl_pre_sum[i + 1][j + 1] = M[i][j] + dagl_pre_sum[i][j]
        
        for i in range(m):
            for j in range(n):
                skew_pre_sum[i + 1][j] = M[i][j] + skew_pre_sum[i][j + 1]

        seen = set()
        heap3 = []
        def dinstinct3(v):
            if v not in seen:
                seen.add(v)
                heapq.heappush(heap3, v)
                if len(heap3) > 3:
                    heapq.heappop(heap3)

        for d in range( (min(m, n) - 1)// 2, 0, -1):
            # 距離 d, element count d + 1
            for i in range(1 + d, 1 + m - d):
                for j in range(d, n - d):
                    # i, j at dagl_pre_sum center's left  
                    # i, j at skew_pre_sum center  
                    a1, b1 = dagl_pre_sum[i + d - 1][j] , dagl_pre_sum[i - 1][j - d]
                    a2, b2 = skew_pre_sum[i + d][j] , skew_pre_sum[i][j + d]
                    a3, b3 = dagl_pre_sum[i][j + 1 + d] , dagl_pre_sum[i - d][j + 1]
                    a4, b4 = skew_pre_sum[i - 1][j + 1- d] , skew_pre_sum[i - 1 - d][j + 1]
                    rhombus = (
                        a1 - b1 +
                        a2 - b2 +
                        a3 - b3 +
                        a4 - b4
                    )

                    dinstinct3(rhombus)

        for row in M:
            for num in row:
                dinstinct3(num)
                
        return heapq.nlargest(3, heap3)



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