<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])

        # link row to col, and link col to row
        graphs = {}  # graphs[v]: the connection graph of value v
        for i in range(m):
            for j in range(n):
                v = matrix[i][j]
                # if not initialized, initial it
                if v not in graphs:
                    graphs[v] = {}
                if i not in graphs[v]:
                    graphs[v][i] = []
                if ~j not in graphs[v]:
                    graphs[v][~j] = []
                # link i to j, and link j to i
                graphs[v][i].append(~j)
                graphs[v][~j].append(i)

        # put points into `value2index` dict, grouped by connection
        value2index = {}  # {v: [[points1], [points2], ...], ...}
        seen = set()  # mark whether put into `value2index` or not

        def dfs(node, graph, rowcols):
            rowcols.add(node)
            for rowcol in graph[node]:
                if rowcol not in rowcols:
                    dfs(rowcol, graph, rowcols)

        for i in range(m):
            for j in range(n):
                if (i, j) in seen:
                    continue
                seen.add((i, j))
                v = matrix[i][j]
                graph = graphs[v]
                # use dfs to find the connected parts
                rowcols = set()   # store visited row and col
                dfs(i, graph, rowcols)
                dfs(~j, graph, rowcols)
                # transform rowcols into points
                points = set()
                for rowcol in rowcols:
                    for k in graph[rowcol]:
                        if k >= 0:
                            points.add((k, ~rowcol))
                            seen.add((k, ~rowcol))
                        else:
                            points.add((rowcol, ~k))
                            seen.add((rowcol, ~k))
                if v not in value2index:
                    value2index[v] = []
                value2index[v].append(points)

        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]:
                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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