Union Find Disjoint Set Data Structure in DSA Typescript - Build from Scratch

Union Find Disjoint Set Data Structure

📖 Scenario: Imagine you have a group of friends, and you want to know which friends are connected directly or through other friends. We will use a Union Find Disjoint Set data structure to keep track of these friend groups.

🎯 Goal: You will build a simple Union Find Disjoint Set structure in TypeScript to group friends and check if two friends are in the same group.

📋 What You'll Learn

Create an array called parent to represent each friend's parent in the group.

Create a variable called size to store the number of friends.

Write a function called find that finds the root parent of a friend.

Write a function called union that connects two friends' groups.

Print the parent array after some unions to show the groups.

💡 Why This Matters

🌍 Real World

Union Find is used in social networks to find friend groups, in computer networks to find connected devices, and in games to manage connected regions.

💼 Career

Understanding Union Find helps in solving problems related to grouping, connectivity, and network design, which are common in software engineering and data science roles.

1

Create the parent array for 5 friends

Create an array called parent with 5 elements where each element is its own parent (0 to 4).

DSA Typescript

// Your code here

Hint

Each friend starts as their own group, so parent[i] should be i.

2

Write the find function

Write a function called find that takes a number x and returns the root parent of x by following the parent array until parent[x] === x.

DSA Typescript

const size = 5;
const parent: number[] = [];
for (let i = 0; i < size; i++) {
    parent[i] = i;
}

// Write the find function below
// Your code here

Hint

Keep moving up the parent until you find the root where parent[x] equals x.

3

Write the union function

Write a function called union that takes two numbers a and b. Use find to get their root parents. If roots are different, set the parent of b's root to a's root to connect the groups.

DSA Typescript

const size = 5;
const parent: number[] = [];
for (let i = 0; i < size; i++) {
    parent[i] = i;
}

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

// Write the union function below
// Your code here

Hint

Connect the root of one group to the root of the other if they are different.

4

Perform unions and print the parent array

Call union(0, 1), union(1, 2), and union(3, 4). Then print the parent array using console.log(parent).

DSA Typescript

const size = 5;
const parent: number[] = [];
for (let i = 0; i < size; i++) {
    parent[i] = i;
}

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

function union(a: number, b: number): void {
    const rootA = find(a);
    const rootB = find(b);
    if (rootA !== rootB) {
        parent[rootB] = rootA;
    }
}

// Perform unions and print the parent array
// Your code here

Hint

After unions, friends 0,1,2 share the same root 0, and friends 3,4 share root 3.

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