<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]]:
        rLen = len(matrix)
        cLen = len(matrix[0])
        
        n = cLen * rLen
        
        rank = [1 for x in range(n)]
        par = [x for x in range(n)]
        
        
        def find(node):
            if par[node] != node:
                par[node] = find(par[node])
            return par[node]
        
        def union(n1, n2):
            p1 = find(n1)
            p2 = find(n2)
            
            if rank[p1] > rank[p2]:
                par[p2] = p1
            elif rank[p2] > rank[p1]:
                par[p1] = p2
            
            else:
                par[p2] = p1
                rank[p1] += 1
        
        
        ans = [[0 for col in range(cLen)] for row in range(rLen)]
        
        row_comps = defaultdict(int)
        col_comps = defaultdict(int)
        
        for row in range(rLen):
            for col in range(cLen):
                uf_index = row*cLen + col
                
                val = matrix[row][col]
                
                if (val, row) in row_comps:
                    union(uf_index, row_comps[(val, row)])
                if (val, col) in col_comps:
                    union(uf_index, col_comps[(val, col)])
                
                row_comps[(val, row)] = uf_index
                col_comps[(val, col)] = uf_index
        
        children = defaultdict(list)
        
        for i in range(n):
            p = find(i)
            children[p].append(i)
        
        topo = []
        for p in children:
            row = p // cLen
            col = p % cLen
            
            topo.append((matrix[row][col], p))
        topo.sort()
        
        rMin = defaultdict(int)
        cMin = defaultdict(int)
        for i, (val, p) in enumerate(topo):
            coords = []
            r = 0
            for child in children[p]:
                row = child // cLen
                col = child % cLen
                r = max(r, max(rMin[row], cMin[col]) + 1)
                coords.append((row, col))

            for row, col in coords:
                ans[row][col] = r
                rMin[row] = r
                cMin[col] = r
        return ans
        
77400