Path Compression in Union Find in DSA Javascript - 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 we use path compression in Union Find.

How does the time to find a set leader change as we do more operations?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the leader of element x and compresses the path to speed up future finds.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to find until the root leader is found.
• How many times: At most once per node on the path from x to the root, but path compression shortens future calls.

How Execution Grows With Input

Initially, the path to the root might be long, so find takes more steps. But each find shortens paths, making future finds faster.

Input Size (n)	Approx. Operations per Find
10	Up to 4-5 steps initially, then close to 1 step
100	Up to 10-15 steps initially, then fewer steps over time
1000	Up to 20-30 steps initially, then almost constant time

Pattern observation: The cost per find decreases over time, approaching almost constant time.

Final Time Complexity

Time Complexity: O(α(n)) (Inverse Ackermann function)

This means each find operation runs almost in constant time for all practical input sizes.

Common Mistake

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

[OK] Correct: Path compression flattens the tree, so after some finds, the height effectively becomes very small, making future finds much faster.

Interview Connect

Understanding path compression's time complexity shows you how clever tweaks make algorithms efficient in practice, a skill valued in problem solving.

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 the Union Find data structure?

easy

A. To speed up the find operation by flattening the tree structure

B. To merge two sets without checking their roots

C. To initialize the parent array with default values

D. To count the number of connected components

Solution

1. Step 1: Understand the role of path compression

   Path compression is used during the find operation to make future queries faster by making nodes point directly to the root.

2. Step 2: Identify the effect on the tree structure

   This flattens the tree, reducing the height and speeding up subsequent find calls.

3. Final Answer:

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

4. Quick Check:

   Path compression = flatten tree = faster find [OK]

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

Common Mistakes:

• Confusing path compression with union by rank
• Thinking it merges sets without root checks
• Believing it initializes data structures

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

easy

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

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

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

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

Solution

1. Step 1: Identify path compression in the code

   function find(x) { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } updates parent[x] to the root during recursion, which is path compression.

2. Step 2: Compare other options

   function find(x) { while (parent[x] !== x) x = parent[x]; return x; } uses iteration but no path compression. function find(x) { if (parent[x] === x) return x; else return find(parent[x]); } uses recursion but no assignment to parent[x]. function find(x) { parent[x] = x; return x; } resets parent incorrectly.

3. Final Answer:

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

4. Quick Check:

   Path compression requires parent update = function find(x) { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } [OK]

Hint: Look for parent update inside find for path compression [OK]

Common Mistakes:

• Missing parent update for path compression
• Using iteration without compression
• Resetting parent incorrectly

3. Given the following 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) {
  if (parent[x] !== x) parent[x] = find(parent[x]);
  return parent[x];
}
find(3);
console.log(parent);

medium

A. [0, 0, 0, 0, 4]

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

C. [0, 0, 1, 1, 4]

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

Solution

1. Step 1: Trace the find(3) call

   find(3) checks parent[3] = 2, not 3, so calls find(2). find(2) checks parent[2] = 1, not 2, so calls find(1). find(1) checks parent[1] = 0, not 1, so calls find(0). find(0) returns 0 since parent[0] = 0.

2. Step 2: Apply path compression updates

   parent[1] = 0, parent[2] = 0, parent[3] = 0 after recursion. But initial array was [0,0,1,2,4]. After updates, parent becomes [0,0,0,0,4]. However, the code only updates parent[x] during recursion. The first call updates parent[3] = find(2) = 0, parent[2] = find(1) = 0, parent[1] = find(0) = 0. So final parent array is [0,0,0,0,4].

3. Final Answer:

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

4. Quick Check:

   Path compression sets parents to root = [0,0,0,0,4] [OK]

Hint: Follow recursion and update parents to root [OK]

Common Mistakes:

• Not updating all parents during recursion
• Confusing parent indices
• Assuming no change after find

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

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

medium

A. Function should use iteration instead of recursion

B. Incorrect base condition for recursion

C. Parent array is not updated for path compression

D. Missing return statement in the else branch

Solution

1. Step 1: Check the recursive calls and returns

   The else branch calls find(parent[x]) but does not return its result, so the function returns undefined for non-root nodes.

2. Step 2: Confirm missing return causes bug

   Without returning the recursive call, the function fails to return the root, breaking the find operation.

3. Final Answer:

   Missing return statement in the else branch -> Option D

4. Quick Check:

   Recursive calls must return values [OK]

Hint: Always return recursive find calls [OK]

Common Mistakes:

• Forgetting to return recursive call result
• Misunderstanding base case
• Thinking iteration is required

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

hard

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

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

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

D. [0, 0, 2, 4, 4]

Solution

1. Step 1: Perform union(3, 4)

   Find roots: find(3) = 2, find(4) = 4. Since rank[2] = 0 and rank[4] = 0, attach 3's root (2) to 4. So parent[2] = 4.

2. Step 2: Perform find(3) with path compression

   find(3) calls find(2), which calls find(4). find(4) returns 4. Path compression sets parent[2] = 4 and parent[3] = parent[2] = 4.

3. Final Answer:

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

4. Quick Check:

   Union attaches root, find compresses path [OK]

Hint: Union attaches root; find compresses path to root [OK]

Common Mistakes:

• Not updating parent after union
• Ignoring path compression updates
• Confusing rank usage