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Union-Find (Disjoint Set Union)
Union-Find is a data structure that tracks elements partitioned into disjoint sets. It supports two operations: union (merge two sets) and find (determine which set an element belongs to).
When to Use
- Problems involving connected components
- Cycle detection in undirected graphs
- Dynamic connectivity queries
- Minimum spanning tree (Kruskal's algorithm)
Common Patterns
Basic Union-Find
Track connected components with path compression
Example: Number of provinces, redundant connection
Weighted Union-Find
Union by rank/size for better balance
Example: Optimize tree depth
Code Templates
Python
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
self.rank = [0] * n
self.count = n # Number of components
def find(self, x):
"""Find with path compression"""
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
"""Union by rank"""
root_x = self.find(x)
root_y = self.find(y)
if root_x == root_y:
return False # Already connected
# Union by rank
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
self.count -= 1
return True
def connected(self, x, y):
"""Check if x and y are in same set"""
return self.find(x) == self.find(y)
def get_count(self):
"""Return number of disjoint sets"""
return self.countC++
class UnionFind {
private:
vector<int> parent;
vector<int> rank;
int count; // Number of components
public:
UnionFind(int n) : parent(n), rank(n, 0), count(n) {
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
// Find with path compression
int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
// Union by rank
bool unite(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) {
return false; // Already connected
}
// Union by rank
if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
} else if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
} else {
parent[rootY] = rootX;
rank[rootX]++;
}
count--;
return true;
}
// Check if x and y are in same set
bool connected(int x, int y) {
return find(x) == find(y);
}
// Return number of disjoint sets
int getCount() {
return count;
}
};Practice Problems
Common Pitfalls
- ⚠️Not implementing path compression (much slower without it)
- ⚠️Forgetting to use find() before comparing roots
- ⚠️Not updating component count after union
- ⚠️Incorrect rank updates in union by rank
Complexity
Time
O(α(n)) amortized per operation, where α is inverse Ackermann
Space
O(n) for parent and rank arrays
Nearly O(1) in practice