<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]]:

        def build_graph(edges):
            graph = [defaultdict(list), defaultdict(list)]
            for x,y in edges:
                graph[0][x].append(y)
                graph[1][y].append(x)
            return graph

        def get_values_pos(matrix, m,n):
            values_pos = {}
            for i in range(m):
                for j in range(n):
                    val = matrix[i][j]
                    values_pos[val] = values_pos.get(val,[])
                    values_pos[val].append((i,j))
            return values_pos

        def get_max_rankdfs(curr, row_or_col, visited, graph, rank):
            curr_rank = 0
            if (curr, row_or_col) not in visited:
                visited.add((curr, row_or_col))
                curr_rank = rank[row_or_col][curr]
                for i in graph[row_or_col][curr]:
                    curr_rank = max(curr_rank, get_max_rankdfs(i, 1^row_or_col, visited, graph, rank))
            return curr_rank
        
                
                


        def solve(matrix):
            m = len(matrix)
            n = len(matrix[0])
            values_pos = get_values_pos(matrix, m,n)
            rank = [[0]*m,[0]*n]
            values = sorted(values_pos.keys())
            for val in values:
                pos = values_pos[val]
                graph = build_graph(pos)
                # print(graph)
                visited = set()
                for x,y in pos:
                    if (x,0) not in visited:
                        nev_visited = set()
                        max_rank = get_max_rankdfs(x,0, nev_visited, graph, rank)
                        for node in nev_visited:
                            visited.add(node)
                            rank[node[1]][node[0]] = max_rank+1
                    matrix[x][y] = rank[0][x]
            return matrix
        
        return solve(matrix)










        
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