<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 typing import List, Tuple
from collections import defaultdict

class Solution:
    def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
        m, n = len(matrix), len(matrix[0])

        # A function to find the root of element x
        def find(x):
            if dsu[x] != x:
                dsu[x] = find(dsu[x])
            return dsu[x]
        
        # Sort the matrix elements along with their coordinates
        elements = []
        for i in range(m):
            for j in range(n):
                elements.append((matrix[i][j], i, j))
        elements.sort()

        # This will store the answer
        answer = [[0] * n for _ in range(m)]
        
        # Ranks to keep track of the maximum rank in each row and column
        rank = [0] * (m + n)

        i = 0
        while i < len(elements):
            # Collect all elements with the same value
            j = i
            value_groups = []
            while j < len(elements) and elements[j][0] == elements[i][0]:
                value_groups.append(elements[j])
                j += 1
            
            # Prepare to union rows and columns
            dsu = list(range(m + n))
            rank2 = rank[:]  # Make a copy of the current ranks
            
            # Merge sets in the union-find
            for _, r, c in value_groups:
                r_root = find(r)
                c_root = find(c + m)
                dsu[r_root] = c_root
                rank2[c_root] = max(rank2[r_root], rank2[c_root])
            
            # Now, assign ranks
            for _, r, c in value_groups:
                root = find(r)
                rank_value = rank2[root] + 1
                answer[r][c] = rank_value
                rank[r] = rank_value
                rank[c + m] = rank_value
            
            # Move to the next group of values
            i = j

        return answer
55100