Union Find
Detailed guide and Python implementation for the 'Union Find' problem.
1. Concept Overview
The 'Union Find' problem is a key challenge in the Graphs section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Union Find.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Union Find carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function detect_cycle_union_find(V, edges) that detects if there is a cycle in an undirected graph of V vertices and a list of edges edges represented as pairs (u, v) using the Union-Find data structure. Return True if a cycle exists, else False.
- •1 <= V <= 1000
- •0 <= len(edges) <= 2000
Examples
V = 3, edges = [(0, 1), (1, 2), (2, 0)]
True
Edge (2, 0) connects vertices 2 and 0 which are already in the same subset, detecting a cycle.
V = 3, edges = [(0, 1), (1, 2)]
False
No subset connections merge vertices already in the same subset.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
class UnionFindOpt:
def __init__(self, n):
self.parent = list(range(n))
self.rank = [0] * n
def find(self, i):
if self.parent[i] == i: return i
self.parent[i] = self.find(self.parent[i])
return self.parent[i]
def union(self, i, j):
root_i = self.find(i)
root_j = self.find(j)
if root_i != root_j:
if self.rank[root_i] < self.rank[root_j]: self.parent[root_i] = root_j
elif self.rank[root_i] > self.rank[root_j]: self.parent[root_j] = root_i
else:
self.parent[root_i] = root_j
self.rank[root_j] += 1Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
class UnionFindBrute:
def __init__(self, n):
self.parent = list(range(n))
def find(self, i):
if self.parent[i] == i: return i
return self.find(self.parent[i])
def union(self, i, j):
self.parent[self.find(i)] = self.find(j)Algorithm Pattern Checklist
When dealing with Graphs data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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