Nothing special here. It’s just a blog post for summarising my algorithm learning course.

# Approach

The Quick Find idea is to assign all the items in 1 connected component with 1 same id. p and q are connected if they have the same id. Every time we do the union command, we update those 2 items with the same id (and also all other items in the connected component).

The set is organised as an array, where the values are the id of that connected component. At first, all the items have the id the same with its index (connected to itself). After each call to union, we update the id of those 2 items to one of them

To check whether the 2 items are connected, simply check whether they have the same id or not. For example

• connected(6, 7) => true because 6 and 7 both have id 1
• connected(0, 9) => false because 0 has the id 1 and 9 has the id 8

# Complexity

Method Complexity
isConnected(p, q) O(1) Only 1 call to check the id of the item
union(p, q) O(n^2) Have to update id of both items all other items having the same id

=> union is too slow for Quick Find.

# Implementation

So easy, no need for sample implementation here. Or did I explain too badly?…