Path Compression in Union Find in DSA Typescript - Time & Space Complexity

Time Complexity: Path Compression in Union Find

O(α(n))

Understanding Time Complexity

We want to understand how fast the "find" operation runs when using path compression in Union Find.

How does the time to find the root change as we do more operations?

Scenario Under Consideration

Analyze the time complexity of the following find function with path compression.

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

This code finds the root of element x and flattens the path to speed up future finds.

Identify Repeating Operations

Look for loops or recursion that repeat work.

• Primary operation: Recursive calls to find following parent links.
• How many times: Up to the height of the tree before compression.

How Execution Grows With Input

Initially, the tree can be tall, so find may take many steps. But path compression flattens it fast.

Input Size (n)	Approx. Operations per find
10	About 2-3 steps after some finds
100	About 4-5 steps initially, fewer after compression
1000	Starts higher but quickly flattens to near 1 step

Pattern observation: The steps grow very slowly and almost stay constant after many finds.

Final Time Complexity

Time Complexity: O(α(n))

This means each find runs in almost constant time, where α(n) is a very slow growing function (the inverse Ackermann function).

Common Mistake

[X] Wrong: "Find always takes time proportional to the height of the tree."

[OK] Correct: Path compression flattens the tree so future finds become much faster, making the average cost almost constant.

Interview Connect

Understanding path compression's time complexity shows you can analyze clever optimizations that make algorithms very efficient in practice.

Self-Check

What if we removed path compression? How would the time complexity of find change?

Practice

1. What is the main purpose of path compression in a Union Find data structure?

easy

A. To sort elements inside each set

B. To speed up the find operation by flattening the tree

C. To delete elements from the set efficiently

D. To merge two sets without checking their roots

Solution

1. Step 1: Understand the find operation

   The find operation locates the root of an element's set in Union Find.

2. Step 2: Role of path compression

   Path compression flattens the tree by making nodes point directly to the root, speeding up future find calls.

3. Final Answer:

   To speed up the find operation by flattening the tree -> Option B

4. Quick Check:

   Path compression = faster find [OK]

Hint: Path compression flattens trees to speed up find [OK]

Common Mistakes:

• Confusing path compression with union operation
• Thinking it deletes elements
• Assuming it sorts elements

2. Which of the following is the correct TypeScript syntax for the find function with path compression in Union Find?

easy

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

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

C. function find(x: number): number { if (parent[x] === x) return x; else return find(parent[x]); }

D. function find(x: number): number { parent[x] = x; return x; }

Solution

1. Step 1: Identify path compression in code

   Path compression updates parent[x] to root during recursion.

2. Step 2: Check each option

   function find(x: number): number { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } updates parent[x] recursively, implementing path compression correctly.

3. Final Answer:

   function find(x: number): number { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } -> Option A

4. Quick Check:

   Path compression updates parent[x] recursively [OK]

Hint: Look for recursive parent update in find [OK]

Common Mistakes:

• Missing parent update for path compression
• Using loop without updating parent
• Returning without compression

3. Given the following TypeScript code snippet, what will be the output of console.log(parent) after calling find(3)?

let parent = [0, 0, 1, 2, 4];
function find(x: number): number {
  if (parent[x] !== x) parent[x] = find(parent[x]);
  return parent[x];
}
find(3);
console.log(parent);

medium

A. [0, 0, 1, 2, 4]

B. [0, 1, 1, 1, 4]

C. [0, 0, 2, 2, 4]

D. [0, 0, 0, 0, 4]

Solution

1. Step 1: Trace find(3) calls

   find(3) checks parent[3] = 2, calls find(2). find(2) checks parent[2] = 1, calls find(1). find(1) checks parent[1] = 0, calls find(0). find(0) returns 0 since parent[0] = 0.

2. Step 2: Apply path compression updates

   Unwinding: parent[1] = 0 (already), returns 0; parent[2] = 0, returns 0; parent[3] = 0, returns 0. Final parent = [0, 0, 0, 0, 4].

3. Final Answer:

   [0, 0, 0, 0, 4] -> Option D

4. Quick Check:

   Path compression sets all path nodes to root [OK]

Hint: Follow recursive calls and update parents on return [OK]

Common Mistakes:

• Not updating parent array after recursion
• Confusing indices and values
• Assuming all parents become root immediately

4. Identify the bug in this TypeScript find function implementing path compression:

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

medium

A. Missing assignment to parent[x] for path compression

B. Incorrect base case condition

C. Function should return void

D. Using recursion without a return statement

Solution

1. Step 1: Analyze recursion and return

   The function calls find(parent[x]) but does not assign its result back to parent[x].

2. Step 2: Identify missing path compression update

   Without parent[x] = find(parent[x]), path compression does not happen, so the tree is not flattened.

3. Final Answer:

   Missing assignment to parent[x] for path compression -> Option A

4. Quick Check:

   Path compression requires updating parent[x] [OK]

Hint: Ensure recursive find result is assigned to parent[x] [OK]

Common Mistakes:

• Forgetting to assign recursive call result
• Returning without recursion
• Wrong base case logic

5. You have a Union Find with parent = [0, 0, 1, 2, 3, 5]. After performing union(4, 5) and then find(5) with path compression, what will be the final parent array?

hard

A. [0, 0, 1, 2, 3, 4]

B. [0, 0, 1, 2, 4, 4]

C. [0, 0, 0, 0, 0, 0]

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

Solution

1. Step 1: Perform union(4, 5)

   Initial parent = [0, 0, 1, 2, 3, 5]. Simple union links by setting parent[5] = 4. Now parent = [0, 0, 1, 2, 3, 4].

2. Step 2: Call find(5) with path compression

   find(5): parent[5]=4 -> find(4): parent[4]=3 -> find(3)=2 -> find(2)=1 -> find(1)=0 -> find(0)=0.

3. Step 3: Update parent array after compression

   Unwinding sets parent[4]=0, parent[3]=0, parent[2]=0, parent[1]=0, but parent[5] remains 4 because find(5) returns 0 but parent[5] was set to 4 and not updated in this code snippet. So after path compression, parent array is [0, 0, 1, 2, 0, 4] if path compression is applied only on the path from 5 to root. However, since find(5) calls find(4), which calls find(3), etc., and path compression updates parent[x] = find(parent[x]) recursively, parent[5] should be updated to 0 as well. So final parent array is [0, 0, 0, 0, 0, 0].

4. Final Answer:

   [0, 0, 0, 0, 0, 0] -> Option C

5. Quick Check:

   Union sets and compress path update parents [OK]

Hint: Union sets then compress path to update parents [OK]

Common Mistakes:

• Not updating parent after union
• Ignoring path compression updates
• Assuming parent stays same after find