Union Find Disjoint Set Data Structure in DSA Typescript

Mental Model

Union Find keeps track of groups of items and quickly tells if two items are in the same group.

Analogy: Imagine friend circles where each person belongs to one circle. Union Find helps you join two circles or check if two people are already friends in the same circle.

0 [parent->0]
1 [parent->1]
2 [parent->2]
3 [parent->3]
4 [parent->4]
Each number points to itself, meaning each is its own group.

Dry Run Walkthrough

Input: Start with 5 elements: 0,1,2,3,4 each in their own group. Perform union(0,1), union(1,2), check if 0 and 2 are connected, then union(3,4), check if 0 and 4 are connected.

Goal: Show how groups merge and how to check if two elements belong to the same group.

Step 1: union(0,1): merge groups containing 0 and 1

0 [parent->0]
1 [parent->0]
2 [parent->2]
3 [parent->3]
4 [parent->4]

Why: We link 1's group to 0's group to combine them.

Step 2: union(1,2): merge groups containing 1 and 2

0 [parent->0]
1 [parent->0]
2 [parent->0]
3 [parent->3]
4 [parent->4]

Why: Since 1's parent is 0, we link 2's group to 0's group, merging 0,1,2.

Step 3: connected(0,2): check if 0 and 2 share the same parent

0 [parent->0]
1 [parent->0]
2 [parent->0]
3 [parent->3]
4 [parent->4]

Why: Both 0 and 2 have parent 0, so they are connected.

Step 4: union(3,4): merge groups containing 3 and 4

0 [parent->0]
1 [parent->0]
2 [parent->0]
3 [parent->3]
4 [parent->3]

Why: We link 4's group to 3's group to combine them.

Step 5: connected(0,4): check if 0 and 4 share the same parent

0 [parent->0]
1 [parent->0]
2 [parent->0]
3 [parent->3]
4 [parent->3]

Why: 0's parent is 0, 4's parent is 3, so they are not connected.

Result:

Groups:
0 -> 1 -> 2 (parent 0)
3 -> 4 (parent 3)
0 and 2 connected: true
0 and 4 connected: false

Annotated Code

DSA Typescript

class UnionFind {
  private parent: number[];
  private rank: number[];

  constructor(size: number) {
    this.parent = Array(size).fill(0).map((_, i) => i);
    this.rank = Array(size).fill(0);
  }

  find(x: number): number {
    if (this.parent[x] !== x) {
      this.parent[x] = this.find(this.parent[x]); // path compression
    }
    return this.parent[x];
  }

  union(x: number, y: number): void {
    const rootX = this.find(x);
    const rootY = this.find(y);
    if (rootX === rootY) return; // already in same group

    if (this.rank[rootX] < this.rank[rootY]) {
      this.parent[rootX] = rootY; // attach smaller tree under bigger
    } else if (this.rank[rootX] > this.rank[rootY]) {
      this.parent[rootY] = rootX;
    } else {
      this.parent[rootY] = rootX;
      this.rank[rootX] += 1; // increase rank if same
    }
  }

  connected(x: number, y: number): boolean {
    return this.find(x) === this.find(y);
  }
}

// Driver code
const uf = new UnionFind(5);
uf.union(0, 1);
uf.union(1, 2);
console.log('0 and 2 connected:', uf.connected(0, 2));
uf.union(3, 4);
console.log('0 and 4 connected:', uf.connected(0, 4));

if (this.parent[x] !== x) { this.parent[x] = this.find(this.parent[x]); // path compression }

compress path to root to speed up future finds

const rootX = this.find(x); const rootY = this.find(y); if (rootX === rootY) return;

find roots and skip union if already connected

if (this.rank[rootX] < this.rank[rootY]) { this.parent[rootX] = rootY; } else if (this.rank[rootX] > this.rank[rootY]) { this.parent[rootY] = rootX; } else { this.parent[rootY] = rootX; this.rank[rootX] += 1; }

attach smaller tree under bigger to keep tree shallow

return this.find(x) === this.find(y);

check if two elements share the same root

Output

0 and 2 connected: true

0 and 4 connected: false

Complexity Analysis

Time: O(α(n)) per operation, where α is the inverse Ackermann function, which is very slow growing and practically constant, because path compression and union by rank keep trees flat.

Space: O(n) for storing parent and rank arrays for n elements.

vs Alternative: Compared to naive union-find without path compression, this approach is much faster for many operations, reducing near-linear time to almost constant.

Edge Cases

union or connected called with same element (e.g., union(2,2))

No change occurs; element is always connected to itself.

DSA Typescript

if (rootX === rootY) return;

checking connected on elements not yet united

Returns false as they belong to different groups.

DSA Typescript

return this.find(x) === this.find(y);

empty UnionFind with zero elements

No operations possible; code handles size 0 gracefully by empty arrays.

DSA Typescript

constructor(size: number) { this.parent = Array(size).fill(0).map((_, i) => i); }

When to Use This Pattern

When you need to quickly group items and check if two items belong to the same group, use Union Find because it efficiently merges groups and checks connectivity.

Common Mistakes

Mistake: Not using path compression in find, causing slow operations.
Fix: Add path compression by setting parent[x] = find(parent[x]) during find.

Mistake: Not using union by rank or size, causing tall trees and slow find.
Fix: Use rank or size to attach smaller tree under bigger tree during union.

Mistake: Forgetting to check if roots are same before union, causing unnecessary changes.
Fix: Check if rootX === rootY before union and skip if true.

Summary

It keeps track of groups of elements and quickly merges or checks if two elements are in the same group.

Use it when you need to manage dynamic connectivity or grouping problems efficiently.

The key is to keep trees flat using path compression and union by rank for fast operations.

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