<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 DSU:
    def __init__(self,n,m):
        self.par=[[(i,j) for j in range(m)] for i in range(n)]
        self.rank=[[1 for j in range(m)] for i in range(n)]
    def find_root(self,a,b):
        if (a,b)==self.par[a][b]: return (a,b)
        pa,pb=self.par[a][b]
        self.par[a][b]=self.find_root(pa,pb)
        return self.par[a][b]
    def union(self,a,b,x,y):
        a,b=self.find_root(a,b)
        x,y=self.find_root(x,y)
        if (a,b)==(x,y): return
        if self.rank[a][b]>self.rank[x][y]:
            self.par[x][y]=(a,b)
            self.rank[a][b]+=self.rank[x][y]
        else:
            self.par[a][b]=(x,y)
            self.rank[x][y]+=self.rank[a][b]

class Solution:
    def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
        n=len(matrix)
        m=len(matrix[0])
        adj=[[[] for j in range(m)] for i in range(n)]
        indeg=[[0 for j in range(m)] for i in range(n)]
        vis=[[False for j in range(m)] for i in range(n)]
        rank=[[1 for j in range(m)] for i in range(n)]
        q=deque()
        dsu=DSU(n,m)
        for i in range(n):
            indices=defaultdict(list)
            for j in range(m): indices[matrix[i][j]].append((i,j))
            for val in indices.values():
                for p0,p1 in zip(val,val[1:]): dsu.union(p0[0],p0[1],p1[0],p1[1])
        for j in range(m):
            indices=defaultdict(list)
            for i in range(n): indices[matrix[i][j]].append((i,j))
            for val in indices.values():
                for p0,p1 in zip(val,val[1:]): dsu.union(p0[0],p0[1],p1[0],p1[1])
        for i in range(n):
            srt=sorted(range(m),key=lambda x: matrix[i][x])
            for j0,j1 in zip(srt,srt[1:]):
                pa,pb=dsu.find_root(i,j0)
                px,py=dsu.find_root(i,j1)
                if (pa,pb)!=(px,py): 
                    adj[pa][pb].append((px,py))
                    indeg[px][py]+=1
        for j in range(m):
            srt=sorted(range(n),key=lambda x: matrix[x][j])
            for i0,i1 in zip(srt,srt[1:]):
                pa,pb=dsu.find_root(i0,j)
                px,py=dsu.find_root(i1,j)
                if (pa,pb)!=(px,py): 
                    adj[pa][pb].append((px,py))
                    indeg[px][py]+=1
        for i in range(n):
            for j in range(m):
                pi,pj=dsu.find_root(i,j)
                if not indeg[pi][pj] and not vis[pi][pj]:
                    vis[pi][pj]=True
                    q.append((pi,pj))
        while q:
            i,j=q.popleft()
            for x,y in adj[i][j]:
                indeg[x][y]-=1
                rank[x][y]=max(rank[x][y],rank[i][j]+1)
                if not indeg[x][y]: q.append((x,y))
        for i in range(n):
            for j in range(m):
                pi,pj=dsu.find_root(i,j)
                rank[i][j]=rank[pi][pj]
        return rank
81400