Union Find Disjoint Set Data Structure in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of the Union Find Disjoint Set data structure?

It helps to keep track of elements divided into groups called sets, and quickly find which set an element belongs to or merge two sets together.

beginner

What does the 'find' operation do in a Union Find Disjoint Set?

It finds the representative or leader of the set that a particular element belongs to, helping to identify if two elements are in the same set.

beginner

What is the purpose of the 'union' operation in Union Find?

It merges two different sets into one set by connecting their leaders, so elements from both sets become part of the same group.

intermediate

Why is path compression used in the 'find' operation?

Path compression flattens the structure of the tree by making each node point directly to the leader, speeding up future 'find' operations.

intermediate

What is the role of 'union by rank' or 'union by size' in Union Find?

It attaches the smaller tree under the root of the larger tree to keep the tree shallow, improving the efficiency of operations.

What does the 'find' operation return in a Union Find data structure?

A The size of the set

B The parent of the element

C The total number of sets

D The leader or representative of the set containing the element

The 'find' operation returns the leader or representative of the set that contains the element.

What is the main benefit of using path compression in Union Find?

A It reduces the height of trees to speed up future finds

B It increases the number of sets

C It merges sets randomly

D It deletes elements from sets

Path compression flattens the tree structure, reducing height and speeding up future 'find' operations.

Which operation merges two sets in Union Find?

A union

B find

C split

D insert

The 'union' operation merges two sets into one.

What does 'union by rank' help to achieve?

A Finds the leader of a set

B Increases the number of sets

C Keeps trees shallow by attaching smaller trees under larger ones

D Deletes duplicate elements

'Union by rank' attaches smaller trees under larger ones to keep the tree shallow and efficient.

If two elements have the same leader, what does it mean?

A They are in different sets

B They belong to the same set

C They have no set

D They are leaders themselves

If two elements share the same leader, they belong to the same set.

Explain how the Union Find Disjoint Set data structure works and why it is useful.

Think about grouping friends and checking if two friends are in the same group.

Describe the optimizations 'path compression' and 'union by rank' in Union Find and their benefits.

Imagine making your friend groups easier to find by shortening paths and keeping groups balanced.

Practice

1. What is the primary purpose of the Union Find Disjoint Set data structure?

easy

A. To efficiently track and merge groups of connected elements

B. To sort elements in ascending order

C. To store key-value pairs for fast lookup

D. To implement a queue with fast enqueue and dequeue

Solution

1. Step 1: Understand the data structure's goal

   The Union Find Disjoint Set is designed to keep track of which elements belong to which groups or sets, especially when these groups merge.

2. Step 2: Compare options with the purpose

   Sorting, key-value storage, and queue operations are unrelated to the main function of Union Find, which is about grouping and merging sets.

3. Final Answer:

   To efficiently track and merge groups of connected elements -> Option A

4. Quick Check:

   Union Find = Track and merge groups [OK]

Hint: Union Find manages connected groups, not sorting or queues [OK]

Common Mistakes:

• Confusing Union Find with sorting algorithms
• Thinking it stores key-value pairs like a dictionary
• Assuming it works like a queue or stack

2. Which of the following is the correct TypeScript syntax to initialize a Union Find class with 5 elements?

easy

A. const uf = UnionFind(5);

B. const uf = new UnionFind(5);

C. let uf = new UnionFind[];

D. let uf = UnionFind.new(5);

Solution

1. Step 1: Recall TypeScript class instantiation syntax

   In TypeScript, to create an instance of a class, we use the new keyword followed by the class name and parentheses with arguments.

2. Step 2: Check each option for correct syntax

   const uf = new UnionFind(5); uses new UnionFind(5); which is correct. const uf = UnionFind(5); misses new. let uf = new UnionFind[]; uses incorrect array syntax. let uf = UnionFind.new(5); uses invalid method call syntax.

3. Final Answer:

   const uf = new UnionFind(5); -> Option B

4. Quick Check:

   Use new ClassName(args) to create instance [OK]

Hint: Use 'new' keyword with class name and parentheses [OK]

Common Mistakes:

• Omitting the 'new' keyword
• Using array syntax instead of class instantiation
• Incorrect method call syntax for constructors

3. Consider the following TypeScript code using a Union Find class:

const uf = new UnionFind(5);
uf.union(0, 1);
uf.union(1, 2);
uf.union(3, 4);
console.log(uf.connected(0, 2));
console.log(uf.connected(0, 4));

What will be the output?

medium

A. true\nfalse

B. false\ntrue

C. true\ntrue

D. false\nfalse

Solution

1. Step 1: Track unions and connections

   Union(0,1) connects 0 and 1. Union(1,2) connects 1 and 2, so 0,1,2 are connected. Union(3,4) connects 3 and 4.

2. Step 2: Check connected queries

   connected(0,2) is true because 0 and 2 are in the same set. connected(0,4) is false because 0 and 4 belong to different sets.

3. Final Answer:

   true false -> Option A

4. Quick Check:

   Connected(0,2) = true, Connected(0,4) = false [OK]

Hint: Track unions step-by-step to check connectivity [OK]

Common Mistakes:

• Assuming all elements are connected initially
• Mixing up union and connected method effects
• Forgetting that union merges sets

4. The following TypeScript code attempts to implement the find method of a Union Find class but has a bug:

find(x: number): number {
  while (this.parent[x] !== x) {
    x = this.parent[x];
  }
  return x;
}

What is the bug and how to fix it?

medium

A. Incorrect comparison; fix by using '==' instead of '!=='

B. Infinite loop possible; fix by adding a break statement

C. Wrong return type; fix by returning boolean instead of number

D. No path compression; fix by setting parent[x] = find(parent[x]) inside loop

Solution

1. Step 1: Identify missing optimization in find method

   The method correctly finds the root but does not apply path compression, which speeds up future queries by flattening the tree.

2. Step 2: Explain fix with path compression

   To fix, recursively find the root and set parent[x] to the root during traversal, e.g., parent[x] = find(parent[x]).

3. Final Answer:

   No path compression; fix by setting parent[x] = find(parent[x]) inside loop -> Option D

4. Quick Check:

   Path compression improves find by updating parents [OK]

Hint: Add path compression to speed up find method [OK]

Common Mistakes:

• Ignoring path compression optimization
• Adding unnecessary break causing incomplete search
• Changing return type incorrectly
• Using wrong equality operator

5. You have 7 elements (0 to 6) initially disconnected. After these union operations:

uf.union(0, 1);
uf.union(1, 2);
uf.union(3, 4);
uf.union(5, 6);
uf.union(4, 5);
uf.union(2, 6);

How many distinct sets remain?

hard

A. 3

B. 2

C. 1

D. 4

Solution

1. Step 1: Track unions step-by-step

   Union(0,1) connects 0 and 1. Union(1,2) connects 2 to 0,1 group. Union(3,4) connects 3 and 4. Union(5,6) connects 5 and 6. Union(4,5) connects sets {3,4} and {5,6} into one group {3,4,5,6}. Union(2,6) connects {0,1,2} and {3,4,5,6} into one big group {0,1,2,3,4,5,6}.

2. Step 2: Count distinct sets after all unions

   All elements are connected into a single set, so only 1 distinct set remains.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   All unions connect all elements into 1 set [OK]

Hint: Follow unions stepwise to count final distinct sets [OK]

Common Mistakes:

• Counting sets before all unions complete
• Missing that union(2,6) merges two big sets
• Assuming initial sets remain unchanged