<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

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

    def find(self, u: int) -> int:
        if self.parent[u] == u:
            return u
        return self.find(self.parent[u])

    def merge(self, u: int, v: int) -> None:
        self.parent.setdefault(u, u)
        self.parent.setdefault(v, v)

        u, v = self.find(u), self.find(v)
        
        if u != v:
            self.parent[u] = v

    def getGroups(self) -> dict[int, list[int]]:
        groups = defaultdict(list)
        for u in self.parent.keys():
            groups[self.find(u)].append(u)
        return groups 

class Solution:
    def matrixRankTransform(self, matrix: list[list[int]]) -> list[list[int]]:
        m, n = len(matrix), len(matrix[0])
        ans = [[0] * n for _ in range(m)]
        positions_dict = defaultdict(list)
        maxRank_list = [0] * (m + n)

        for i, row in enumerate(matrix):
            for j, val in enumerate(row):
                positions_dict[val].append((i, j))

        for _, positions in sorted(positions_dict.items()):
            nds = NaiveDisjointSet()

            for i, j in positions:
                nds.merge(i, j+m)

            for values in nds.getGroups().values():
                maxRank = max(maxRank_list[i] for i in values)
                for i in values:
                    maxRank_list[i] = maxRank + 1

            for i, j in positions:
                ans[i][j] = maxRank_list[i]

        return ans
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