Bird
Raised Fist0
DSA Typescriptprogramming~5 mins

Union Find Disjoint Set Data Structure in DSA Typescript - Time & Space Complexity

Choose your learning style10 modes available
Time Complexity: Union Find Disjoint Set Data Structure
O(α(n))
Understanding Time Complexity

We want to understand how fast the Union Find data structure works when connecting groups and checking if elements belong together.

How does the time needed change as we add more elements and connections?

Scenario Under Consideration

Analyze the time complexity of the following Union Find operations.


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 groups of connected elements and finds which group an element belongs to, using path compression.

Identify Repeating Operations

Look at the main repeated actions:

  • Primary operation: The find method, which follows parent links up to the root.
  • How many times: Each union calls find twice; many find calls happen during unions and checks.
How Execution Grows With Input

As we add more elements and unions, the time per operation grows very slowly.

Input Size (n)Approx. Operations per find/union
10About 4 steps
100About 5 steps
1000About 6 steps

Pattern observation: The steps grow very slowly, almost like a flat line, thanks to path compression.

Final Time Complexity

Time Complexity: O(α(n)) per operation

This means each find or union runs in almost constant time, where α(n) is a very slow-growing function.

Common Mistake

[X] Wrong: "Each find operation takes linear time because it may follow many parents."

[OK] Correct: Path compression flattens the structure, so repeated finds become very fast, not linear.

Interview Connect

Understanding Union Find's time complexity shows you can handle efficient grouping problems, a common skill in coding challenges.

Self-Check

"What if we removed path compression from the find method? How would the time complexity change?"

Practice

(1/5)
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
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard