Introduction

In this blog post, we will explore how to solve the LeetCode problem "1632" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. The test cases are generated so 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: 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]] Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 500 -109 <= matrix[row][col] <= 109

Explanation

Here's the breakdown of the solution:

• Union-Find for Equality: Use the Union-Find data structure to group elements that are equal within the same row or column. This helps identify elements that must have the same rank.
• Topological Sort (Implicit): Process elements in ascending order of their values. For each element, determine its rank based on the maximum rank of its neighbors (in the same row or column) that have smaller values. This can be viewed as an implicit topological sort because we are essentially processing nodes (matrix elements) in an order such that all dependencies (smaller ranked neighbors) are processed first.

• Rank Propagation: After calculating the initial rank for an element, propagate this rank to all other elements in its Union-Find group, ensuring that equal elements have the same rank.

• Runtime Complexity: O(m*n*alpha(m*n)), where alpha is the inverse Ackermann function (effectively constant).

• Storage Complexity: O(m*n)

Code

    class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n

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

def matrix_rank_transform(matrix):
    m, n = len(matrix), len(matrix[0])
    result = [[0] * n for _ in range(m)]
    elements = []
    for r in range(m):
        for c in range(n):
            elements.append((matrix[r][c], r, c))

    elements.sort()

    row_rank = [0] * m
    col_rank = [0] * n

    i = 0
    while i < len(elements):
        j = i
        group = []
        while j < len(elements) and elements[i][0] == elements[j][0]:
            group.append(elements[j])
            j += 1

        row_uf = UnionFind(m)
        col_uf = UnionFind(n)

        for _, r, c in group:
            row_uf.union(r, r)
            col_uf.union(c, c)

        row_max_rank = {}
        col_max_rank = {}

        for _, r, c in group:
            row_root = row_uf.find(r)
            col_root = col_uf.find(c)
            if row_root not in row_max_rank:
                row_max_rank[row_root] = row_rank[r]
            if col_root not in col_max_rank:
                col_max_rank[col_root] = col_rank[c]

        for _, r, c in group:
            row_root = row_uf.find(r)
            col_root = col_uf.find(c)
            rank = max(row_max_rank[row_root], col_max_rank[col_root]) + 1
            result[r][c] = rank

        for _, r, c in group:
            row_root = row_uf.find(r)
            col_root = col_uf.find(c)
            row_rank[r] = max(row_rank[r], result[r][c])
            col_rank[c] = max(col_rank[c], result[r][c])

        i = j

    return result