<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>

from collections import defaultdict
from typing import List, Dict, Tuple

class DisjointSet:
    def __init__(self):
        self.parent = {}

    def find_root(self, x: int) -> int:
        if self.parent.get(x, x) == x:
            return x
        self.parent[x] = self.find_root(self.parent[x])
        return self.parent[x]

    def unite(self, x: int, y: int) -> None:
        self.parent.setdefault(x, x)
        self.parent.setdefault(y, y)
        root_x = self.find_root(x)
        root_y = self.find_root(y)
        if root_x != root_y:
            self.parent[root_x] = root_y

    def get_components(self) -> Dict[int, List[int]]:
        component_map = defaultdict(list)
        for x in self.parent.keys():
            root_x = self.find_root(x)
            component_map[root_x].append(x)
        return component_map

class Solution:
    def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
        rows = len(matrix)
        cols = len(matrix[0])
        result = [[0] * cols for _ in range(rows)]

        # {value: [(i, j)]}
        value_to_positions = defaultdict(list)
        # ranks[i] := the maximum rank of the row or column so far
        current_rank = [0] * (rows + cols)

        for i in range(rows):
            for j in range(cols):
                value = matrix[i][j]
                value_to_positions[value].append((i, j))

        for value, positions in sorted(value_to_positions.items()):
            ds = DisjointSet()
            for i, j in positions:
                # Unite i-th row with j-th column
                ds.unite(i, j + rows)

            for component in ds.get_components().values():
                # Get the maximum rank of all the included rows and columns
                max_rank = max(current_rank[i] for i in component)
                # Update all the rows and columns to max_rank + 1
                for i in component:
                    current_rank[i] = max_rank + 1

            for i, j in positions:
                result[i][j] = current_rank[i]

        return result

55800