Path Compression in Union Find in DSA C - 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.

Specifically, how does the time grow as we have more elements and unions?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the root parent of element x and compresses the path by making x point directly to the root.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to find until the root is reached.
• How many times: At most the height of the tree representing the set.

How Execution Grows With Input

Initially, trees can be tall, so find may take longer. But path compression flattens trees over time.

Input Size (n)	Approx. Operations per find
10	About 2-3 steps after some finds
100	About 4-5 steps after many finds
1000	About 5-6 steps after many finds

Pattern observation: The number of steps grows very slowly, almost constant for practical sizes.

Final Time Complexity

Time Complexity: O(α(n))

This means each find runs in almost constant time, where α(n) is the inverse Ackermann function, which grows extremely slowly.

Common Mistake

[X] Wrong: "Path compression makes find run in constant time always."

[OK] Correct: Initially, find can take longer if trees are tall, but path compression quickly flattens them, making future finds very fast.

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 initialize the parent array

B. To merge two sets by size

C. To count the number of connected components

D. To speed up the find operation by making nodes point directly to the root

Solution

1. Step 1: Understand the find operation

   The find operation locates the root parent of a node in Union Find.

2. Step 2: Role of path compression

   Path compression makes all nodes on the path point directly to the root, speeding up future find calls.

3. Final Answer:

   To speed up the find operation by making nodes point directly to the root -> Option D

4. Quick Check:

   Path compression optimizes find = C [OK]

Hint: Path compression flattens tree for faster find [OK]

Common Mistakes:

• Confusing path compression with union by size
• Thinking it initializes the structure
• Believing it counts components

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

easy

A. int find(int x) { while(parent[x] != x) x = parent[x]; return x; }

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

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

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

Solution

1. Step 1: Identify path compression in code

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

2. Step 2: Check each option

   int find(int x) { if(parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } updates parent[x] with recursive find result, implementing path compression correctly.

3. Final Answer:

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

4. Quick Check:

   Path compression updates parent recursively = A [OK]

Hint: Look for parent update inside find recursion [OK]

Common Mistakes:

• Using loop without updating parent (Option B)
• Not updating parent during recursion (Option C)
• Resetting parent incorrectly (Option D)

3. Given the following Union Find structure with parent = [0, 0, 1, 2, 3], what will be the parent array after calling find(4) with path compression?

medium

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

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

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

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

Solution

1. Step 1: Trace find(4) path

   4's parent is 3, 3's parent is 2, 2's parent is 1, 1's parent is 0, which is root.

2. Step 2: Apply path compression

   All nodes on path (4,3,2,1) update parent to root 0 (parent[1] already 0). parent[2]=0, parent[3]=0, parent[4]=0. So parent array becomes [0,0,0,0,0].

3. Final Answer:

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

4. Quick Check:

   Path compression sets parent[4] = 0 = A [OK]

Hint: Update all parents on path to root after find [OK]

Common Mistakes:

• Only updating parent[4] to root (Option B)
• Not updating any parent (Option C)
• Updating wrong parents (Option D)

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

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

medium

A. No bug, code is correct

B. Incorrect base case condition

C. Missing return value from recursive call before assignment

D. Parent update should be after return

Solution

1. Step 1: Analyze recursive call

   The recursive call find(parent[x]) is made but its return value is ignored.

2. Step 2: Check parent update

   Parent update uses parent[parent[x]] instead of the root returned by recursion, so path compression fails.

3. Final Answer:

   Missing return value from recursive call before assignment -> Option C

4. Quick Check:

   Return value must be assigned to parent[x] = D [OK]

Hint: Assign recursive find result to parent[x] for compression [OK]

Common Mistakes:

• Ignoring recursive return value
• Updating parent incorrectly
• Misunderstanding base case

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

hard

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

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

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

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

Solution

1. Step 1: Perform union(4, 0)

   Find roots: root(4) = 4 (chain 4->3->2->1->0), root(0) = 0. Since roots differ, link root(4) to root(0). So parent[4] = 0. Updated parent = [0, 0, 1, 2, 0].

2. Step 2: Call find(4) with path compression

   Path: 4 -> 0. Since 0 is root, parent[4] already 0, no further updates. Final parent = [0, 0, 1, 2, 0].

3. Final Answer:

   [0, 0, 1, 2, 0] -> Option A

4. Quick Check:

   Union links 4 to 0, find compresses path = A [OK]

Hint: Union links roots, find compresses path to root [OK]

Common Mistakes:

• Incorrectly linking assuming same root (Option C)
• Wrong union target (Option B)
• Incomplete path compression (Option D)