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Problem solution · Python

Check if the Rectangle Corner Is Reachable

Check if the Rectangle Corner Is Reachable: a Python solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Check if the Rectangle Corner Is Reachable, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 52 lines of Python from the credited upstream file 3235.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCheck if the Rectangle Corner Is Reachable · PythonPython
Use this to learn the idea, then write your own version.
class UnionFind:  def __init__(self, n: int):    self.id = list(range(n))    self.rank = [0] * n   def unionByRank(self, u: int, v: int) -> None:    i = self.find(u)    j = self.find(v)    if i == j:      return    if self.rank[i] < self.rank[j]:      self.id[i] = j    elif self.rank[i] > self.rank[j]:      self.id[j] = i    else:      self.id[i] = j      self.rank[j] += 1   def find(self, u: int) -> int:    if self.id[u] != u:      self.id[u] = self.find(self.id[u])    return self.id[u]  class Solution:  def canReachCorner(self, X: int, Y: int, circles: list[list[int]]) -> bool:    n = len(circles)    # Add two virtual nodes, where node n represents (0, 0) and node n + 1    # represents (X, Y).    uf = UnionFind(n + 2)     # Iterate through each circle.    for i, (x, y, r) in enumerate(circles):      # Union the current circle with the node (0, 0) if the circle overlaps      # with the left or top edges.      if x - r <= 0 or y + r >= Y:        uf.unionByRank(i, n)      # Union the current circle with the node (X, Y) if the circle overlaps      # with the right or bottom edges.      if x + r >= X or y - r <= 0:        uf.unionByRank(i, n + 1)      # Union the current circle with previous circles if they overlap.      for j in range(i):        x2, y2, r2 = circles[j]        if (x - x2)**2 + (y - y2)**2 <= (r + r2)**2:          uf.unionByRank(i, j)     # If nodes (0, 0) and (X, Y) are in the same union set, that means there's    # a path of overlapping circles that connects the left or top edges to the    # right or bottom edges, implying that (0, 0) cannot reach (X, Y).    return uf.find(n) != uf.find(n + 1)

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