<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]]:
        import collections
        import heapq

        m, n = len(matrix), len(matrix[0])
        rank = [0] * (m + n)  # rank[i] represents the max rank for row i and column i+m

        # Group elements by their values
        val_to_pos = collections.defaultdict(list)
        for i in range(m):
            for j in range(n):
                val_to_pos[matrix[i][j]].append((i, j))

        # Union-Find data structure
        class UnionFind:
            def __init__(self, size):
                self.parent = list(range(size))
                self.rank = [0] * size

            def find(self, x):
                if self.parent[x] != x:
                    self.parent[x] = self.find(self.parent[x])
                return self.parent[x]

            def union(self, x, y):
                root_x, root_y = self.find(x), self.find(y)
                if root_x != root_y:
                    if self.rank[root_x] > self.rank[root_y]:
                        self.parent[root_y] = root_x
                    elif self.rank[root_x] < self.rank[root_y]:
                        self.parent[root_x] = root_y
                    else:
                        self.parent[root_y] = root_x
                        self.rank[root_x] += 1

        # Process elements in increasing order
        result = [[0] * n for _ in range(m)]
        for val in sorted(val_to_pos):
            uf = UnionFind(m + n)
            rank2 = rank[:]
            for i, j in val_to_pos[val]:
                root_i = uf.find(i)
                root_j = uf.find(j + m)
                uf.union(i, j + m)
                root = uf.find(i)
                rank2[root] = max(rank2[root_i], rank2[root_j])
            for i, j in val_to_pos[val]:
                rank[i] = rank[j + m] = rank2[uf.find(i)] + 1
                result[i][j] = rank[i]

        return result

# Example usage:
solution = Solution()
matrix = [[1, 2], [3, 4]]
print(solution.matrixRankTransform(matrix))  # Output: [[1, 2], [2, 3]]
55900