# Given an m x n matrix, return a new matrix answer where answer[row][col] is the rank of matrix[row][col].
#
# The rank is an integer that represents how large an element is compared to other elements. It is calculated using the following rules:
#
#
# 	The rank is an integer starting from 1.
# 	If two elements p and q are in the same row or column, then:
#
# 		If p < q then rank(p) < rank(q)
# 		If p == q then rank(p) == rank(q)
# 		If p > q then rank(p) > rank(q)
#
#
# 	The rank should be as small as possible.
#
#
# It is guaranteed that answer is unique under the given rules.
#
#  
# Example 1:
#
#
# Input: matrix = [[1,2],[3,4]]
# Output: [[1,2],[2,3]]
# Explanation:
# 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] > matrix[0][0] and matrix[0][0] is rank 1.
# The rank of matrix[1][0] is 2 because matrix[1][0] > matrix[0][0] and matrix[0][0] is rank 1.
# The rank of matrix[1][1] is 3 because matrix[1][1] > matrix[0][1], matrix[1][1] > matrix[1][0], and both matrix[0][1] and matrix[1][0] are rank 2.
#
#
# Example 2:
#
#
# Input: matrix = [[7,7],[7,7]]
# Output: [[1,1],[1,1]]
#
#
# Example 3:
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#
# Input: matrix = [[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]
# Output: [[4,2,3],[1,3,4],[5,1,6],[1,3,4]]
#
#
# Example 4:
#
#
# Input: matrix = [[7,3,6],[1,4,5],[9,8,2]]
# Output: [[5,1,4],[1,2,3],[6,3,1]]
#
#
#  
# Constraints:
#
#
# 	m == matrix.length
# 	n == matrix[i].length
# 	1 <= m, n <= 500
# 	-109 <= matrix[row][col] <= 109
#
#


class Solution:
    def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
        cm = defaultdict(list)
        n = len(matrix)
        m = len(matrix[0])
        for i in range(n):
            for j in range(m): 
                # use cm to record all occurences value in the matrix
                cm[matrix[i][j]].append([i, j])
                
        def find(i):
            if p[i]<0: 
                return i
            else: 
                p[i] = find(p[i])
                return p[i]

        # keep track of maximum ranks used for lines and columns 
        ranks = [0 for _ in range(n+m)]
        
        for a in sorted(cm):
            p = [-1 for _ in range(n+m)]
            nxt_rank = ranks[:]
            for i, j in cm[a]:
                i, j = find(i), find(j + n)
                if (i != j):
                    p[j] = i
                nxt_rank[i] = max(nxt_rank[i], nxt_rank[j])
            for i, j in cm[a]:
                ranks[i] = ranks[j + n] = matrix[i][j] = nxt_rank[find(i)] + 1
        
        return matrix
