Leetcode #1632: Rank Transform of a Matrix
In this guide, we solve Leetcode #1632 Rank Transform of a Matrix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Graph, Topological Sort, Array, Matrix, Sorting
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa != pb:
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
def reset(self, x):
self.p[x] = x
self.size[x] = 1
class Solution:
def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
m, n = len(matrix), len(matrix[0])
d = defaultdict(list)
for i, row in enumerate(matrix):
for j, v in enumerate(row):
d[v].append((i, j))
row_max = [0] * m
col_max = [0] * n
ans = [[0] * n for _ in range(m)]
uf = UnionFind(m + n)
for v in sorted(d):
rank = defaultdict(int)
for i, j in d[v]:
uf.union(i, j + m)
for i, j in d[v]:
rank[uf.find(i)] = max(rank[uf.find(i)], row_max[i], col_max[j])
for i, j in d[v]:
ans[i][j] = row_max[i] = col_max[j] = 1 + rank[uf.find(i)]
for i, j in d[v]:
uf.reset(i)
uf.reset(j + m)
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.