# Union Find

### Remarks

- Union Find is a data structure keeps track of a set of elements that are partitioned into a number of disjoint subsets.
- It has two operations
`union(p, q)`

and`find(p)`

.- the
**find/search**finds the subset element`p`

belongs to. - and
**union/merge**merges the subsets containing`p`

and`q`

.

- the
- It memory usage is O(N), and each
**find**and**union**operations are near O(1) in time. - The algorithm can be used to find all connected components in a network.
- It is also used in kruskal’s algorithm to find the minimal spanning tree for a graph.

### Implementation

Simplest implementation to use the core of union find.

```
parents = list(range(n))
def find(i):
if i != parents[i]:
parents[i] = find(parents[i])
i = parents[i]
return parents[i]
def union(x, y):
x, y = find(x), find(y)
if x == y: return
parents[x] = y
return y
```

Define a class for UF

```
class UnionFind:
def __init__(self, n):
self.parents = list(range(n))
self.sizes = [1] * n
def _find_recursive(self, i):
while i != self.parents[i]:
# path compression, have i points to the cluster centroid
self.parents[i] = self._find_recursive(self.parents[i])
i = self.parents[i]
return i
def _find_iterative(self, i):
# 1 path to find the root
root = i
while self.parents[root] != root:
root = self.parents[root]
# 2 path to assign every node in the path to points at root
while self.parents[i] != root:
parent = self.parents[i]
self.parents[i] = root
i = parent
return root
find = _find_recursive
def union(self, p, q):
"""Attempt to union two elements
Args:
p, q: two elements to be attempted for union
Returns:
Boolean indicates whether a union operation is performed.
"""
root_p, root_q = map(self.find, (p, q))
if root_p == root_q: return False
small, big = sorted([root_p, root_q], key=lambda x: self.sizes[x])
self.parents[small] = big
self.sizes[big] += self.sizes[small]
return True
```

Use Union Find to find connected components in undirected graph.

```
connections = [[0, 1], [1, 2], [2, 3], [3, 4], [5, 6], [6, 8], [7, 9]]
uf = UnionFind(10)
for p, q in connections: uf.union(p, q)
num_components = len(set(uf.find(i) for i in range(10)))
print(num_components)
```

- To find minimum spanning tree in the graph where there are weights on the edges.
- Sort edges by edge weights ascendingly.
- Iteratve over the edges unify nodes when two nodes don’t belong to same cluster.
- Repeat 2 until no nodes or no edges.

### An Implementation without pre-allocation

```
class UnionFind:
def __init__(self):
self.parents = dict()
self.sizes = dict()
self.n_sets = 0
def __contains__(self, i):
return i in self.parents
def insert(self, i):
if self.__contains__(i): return
self.parents[i] = i
self.sizes[i] = 1
self.n_sets += 1
def find(self, i):
while i != self.parents[i]:
self.parents[i] = self.find(self.parents[i])
i = self.parents[i]
return i
def union(self, p, q):
"""Attempt to union two elements
Returns:
Boolean indicates whether a union operation is performed.
"""
root_p, root_q = map(self.find, (p, q))
if root_p == root_q: return False
small, big = sorted([root_p, root_q], key=lambda x: self.sizes[x])
self.parents[small] = big
self.sizes[big] += self.sizes[small]
self.n_sets -= 1
return True
```