<p>Given an <code>m x n</code> <code>matrix</code>, return <em>a new matrix </em><code>answer</code><em> where </em><code>answer[row][col]</code><em> is the </em><em><strong>rank</strong> of </em><code>matrix[row][col]</code>.</p>

<p>The <strong>rank</strong> is an <strong>integer</strong> that represents how large an element is compared to other elements. It is calculated using the following rules:</p>

<ul>
	<li>The rank is an integer starting from <code>1</code>.</li>
	<li>If two elements <code>p</code> and <code>q</code> are in the <strong>same row or column</strong>, then:
	<ul>
		<li>If <code>p &lt; q</code> then <code>rank(p) &lt; rank(q)</code></li>
		<li>If <code>p == q</code> then <code>rank(p) == rank(q)</code></li>
		<li>If <code>p &gt; q</code> then <code>rank(p) &gt; rank(q)</code></li>
	</ul>
	</li>
	<li>The <strong>rank</strong> should be as <strong>small</strong> as possible.</li>
</ul>

<p>The test cases are generated so that <code>answer</code> is unique under the given rules.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/18/rank1.jpg" style="width: 442px; height: 162px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2],[3,4]]
<strong>Output:</strong> [[1,2],[2,3]]
<strong>Explanation:</strong>
The rank of matrix[0][0] is 1 because it is the smallest integer in its row and column.
The rank of matrix[0][1] is 2 because matrix[0][1] &gt; matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][0] is 2 because matrix[1][0] &gt; matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][1] is 3 because matrix[1][1] &gt; matrix[0][1], matrix[1][1] &gt; matrix[1][0], and both matrix[0][1] and matrix[1][0] are rank 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/18/rank2.jpg" style="width: 442px; height: 162px;" />
<pre>
<strong>Input:</strong> matrix = [[7,7],[7,7]]
<strong>Output:</strong> [[1,1],[1,1]]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/18/rank3.jpg" style="width: 601px; height: 322px;" />
<pre>
<strong>Input:</strong> matrix = [[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]
<strong>Output:</strong> [[4,2,3],[1,3,4],[5,1,6],[1,3,4]]
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>-10<sup>9</sup> &lt;= matrix[row][col] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def matrixRankTransform(self, M: List[List[int]]) -> List[List[int]]:
        mr = [0 for _ in M]
        mc = [0 for _ in range(len(M[0]))]
        ii = sorted([(i, j) for j in range(len(M[0])) for i in range(len(M))], key=lambda x: M[x[0]][x[1]])
        rr = [[0 for _ in r] for r in M]
        for _, gg in groupby(ii, lambda x: M[x[0]][x[1]]):
            gg = list(gg)
            pp = [jj for jj in range(len(gg))]
            ra = [1 for _ in range(len(gg))]
            def find(a: int) -> int:
                if a != pp[a]:
                    pp[a] = find(pp[a])
                return pp[a]
            def join(a: int, b: int):
                pa = find(a)
                pb = find(b)
                if pa == pb: return
                if ra[pa] > ra[pb]:
                    pp[pb] = pa
                    ra[pa] += ra[pb]
                else:
                    pp[pa] = pb
                    ra[pb] += ra[pa]
            
            rd, cd = {}, {}
            for aa in range(len(gg)):
                join(aa, rd.get(gg[aa][0], aa))
                join(aa, cd.get(gg[aa][1], aa))
                rd[gg[aa][0]] = aa
                cd[gg[aa][1]] = aa
            
            mnn = defaultdict(int)
            for aa in range(len(gg)):
                p = find(aa)
                mnn[p] = max(mnn[p], max(mr[gg[aa][0]], mc[gg[aa][1]]) + 1)
            
            for aa in range(len(gg)):
                mn = mnn[find(aa)]
                rr[gg[aa][0]][gg[aa][1]] = mn
                mr[gg[aa][0]] = mn
                mc[gg[aa][1]] = mn
            
        return rr
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