<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, matrix: List[List[int]]) -> List[List[int]]:
        m = len(matrix)
        n = len(matrix[0])

        # implement find and union
        def find(UF, x):
            if x != UF[x]:
                UF[x] = find(UF, UF[x])
            return UF[x]

        def union(UF, x, y):
            UF.setdefault(x, x)
            UF.setdefault(y, y)
            UF[find(UF, x)] = find(UF, y)

        # link row and col together
        UFs = {}  # UFs[v]: the Union-Find of value v
        for i in range(m):
            for j in range(n):
                v = matrix[i][j]
                if v not in UFs:
                    UFs[v] = {}
                # union i to j
                union(UFs[v], i, ~j)

        # put points into `value2index` dict, grouped by connection
        value2index = {}
        for i in range(m):
            for j in range(n):
                v = matrix[i][j]
                if v not in value2index:
                    value2index[v] = {}
                f = find(UFs[v], i)
                if f not in value2index[v]:
                    value2index[v][f] = []
                value2index[v][f].append((i, j))

        answer = [[0]*n for _ in range(m)]  # the required rank matrix
        rowmax = [0] * m  # rowmax[i]: the max rank in i row
        colmax = [0] * n  # colmax[j]: the max rank in j col
        for v in sorted(value2index.keys()):
            # update by connected points with same value
            for points in value2index[v].values():
                rank = 1
                for i, j in points:
                    rank = max(rank, max(rowmax[i], colmax[j]) + 1)
                for i, j in points:
                    answer[i][j] = rank
                    # update rowmax and colmax
                    rowmax[i] = max(rowmax[i], rank)
                    colmax[j] = max(colmax[j], rank)

        return answer

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