Union Find Disjoint Set Data Structure in DSA Typescript - Execution Trace

Concept Flow - Union Find Disjoint Set Data Structure

Initialize each element as its own set

↓

Find operation: Find root of element

↓

Union operation: Merge two sets

↓

Update parent pointers

↓

Repeat for queries

↓

Sets merged

Start with each element alone. Find finds the root parent. Union merges sets by linking roots. Repeat to combine sets.

Execution Sample

DSA Typescript

class UnionFind {
  parent: number[];
  constructor(n: number) {
    this.parent = Array.from({ length: n }, (_, i) => i);
  }
  find(x: number): number {
    if (this.parent[x] !== x) this.parent[x] = this.find(this.parent[x]);
    return this.parent[x];
  }
  union(a: number, b: number): void {
    const rootA = this.find(a);
    const rootB = this.find(b);
    if (rootA !== rootB) this.parent[rootB] = rootA;
  }
}

This code creates sets for elements 0 to n-1, finds roots with path compression, and unions sets by linking roots.

Execution Table

Step	Operation	Elements Involved	Parent Array State	Pointer Changes	Visual State
1	Initialize sets	0,1,2,3,4	[0,1,2,3,4]	Each element points to itself	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 1 │ │ = 2 │ │ = 3 │ │ = 4 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
2	Union(0,1)	0,1	[0,0,2,3,4]	parent[1] = 0	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 0 │ │ = 2 │ │ = 3 │ │ = 4 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
3	Union(3,4)	3,4	[0,0,2,3,3]	parent[4] = 3	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 0 │ │ = 2 │ │ = 3 │ │ = 3 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
4	Union(1,3)	1,3	[0,0,2,0,3]	parent[3] = 0 (after find(1)=0, find(3)=3)	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 0 │ │ = 2 │ │ = 0 │ │ = 3 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
5	Find(4)	4	[0,0,2,0,0]	Path compression: parent[4] = find(3) = 0	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 0 │ │ = 2 │ │ = 0 │ │ = 0 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
6	Check if 2 and 4 connected	2,4	[0,0,2,0,0]	find(2)=2, find(4)=0, different roots	Sets are separate: {0,1,3,4} and {2}
7	Union(4,2)	2,4	[0,0,0,0,0]	parent[2] = 0	All elements connected: ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:0 │ │ data:1 │ │ data:2 │ │ data:3 │ │ data:4 │ │ parent │ │ parent │ │ parent │ │ parent │ │ parent │ │ = 0 │ │ = 0 │ │ = 0 │ │ = 0 │ │ = 0 │ └────────┘ └────────┘ └────────┘ └────────┘ └────────┘
8	Find(2)	2	[0,0,0,0,0]	find(2) = 0 (root)	No pointer changes (already compressed)
9	End	-	[0,0,0,0,0]	-	All elements in one set with root 0

💡 All unions processed; final parent array shows all elements connected under root 0

Variable Tracker

Variable	Start	After 2	After 3	After 4	After 5	After 7	Final
parent	[0,1,2,3,4]	[0,0,2,3,4]	[0,0,2,3,3]	[0,0,2,0,3]	[0,0,2,0,0]	[0,0,0,0,0]	[0,0,0,0,0]

Key Moments - 3 Insights

Why does the parent array change during find(4) in step 5?

In step 5, find(4) triggers path compression. The parent of 4 changes from 3 to 0 to speed up future finds. See execution_table row 5 for pointer change.

Why do we only update parent of rootB in union, not both?

Union links rootB's parent to rootA to merge sets without cycles. Updating both would break tree structure. See step 4 and 7 in execution_table.

Why is find used before union?

Find gets the root parent to know which sets elements belong to. Union merges only if roots differ, avoiding redundant merges. See steps 2,3,4 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the parent of element 4 after find(4)?

A 3

B 0

C 4

D 1

💡 Hint

Check the 'Pointer Changes' and 'Parent Array State' columns in step 5.

At which step does the parent of element 3 change to 0?

A Step 4

B Step 6

C Step 2

D Step 7

💡 Hint

Look at the 'Pointer Changes' column and 'Parent Array State' in step 4.

If we skip path compression in find, how would the parent array after step 5 differ?

A [0,0,2,3,3]

B [0,0,0,0,0]

C [0,0,2,0,3]

D [0,1,2,3,4]

💡 Hint

Compare step 4 and 5 parent arrays; path compression updates parent[4].

Concept Snapshot

Union Find Disjoint Set:
- Initialize each element as its own parent
- find(x): returns root parent with path compression
- union(a,b): link roots if different
- Efficiently tracks connected components
- Parent array shows set leaders
- Path compression flattens tree for speed

Full Transcript

Union Find Disjoint Set starts with each element in its own set. The find operation locates the root parent of an element, compressing paths to speed future queries. The union operation merges two sets by linking the root of one to the root of another if they differ. The parent array tracks these links. Steps show initializing sets, performing unions, and path compression during finds. Key moments clarify why parent pointers update during find and why union only updates one root. Visual quizzes test understanding of parent changes and path compression effects.

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