Path Compression in Union Find in DSA Java - 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 x) {
  if (parent[x] != x) {
    parent[x] = find(parent[x]);
  }
  return parent[x];
}
    

This code finds the root 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 reaching the root.
• How many times: At most once per node on the path from x to root, but path compression shortens future calls.

How Execution Grows With Input

Initially, paths can be long, so find may take several steps. But path compression flattens the tree, making future finds faster.

Input Size (n)	Approx. Operations per find
10	About 2-3 steps on average
100	About 4-5 steps initially, fewer after compression
1000	About 5-6 steps initially, very few after compression

Pattern observation: The number of steps grows very slowly and flattens quickly due to path compression.

Final Time Complexity

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

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

Common Mistake

[X] Wrong: "Find with path compression is O(n) because it uses recursion and can visit many nodes."

[OK] Correct: Path compression flattens the tree so future finds are much faster, making the average cost almost constant, not linear.

Interview Connect

Understanding path compression's time complexity shows you know how to make Union Find efficient, a key skill in many graph and connectivity problems.

Self-Check

"What if we removed path compression and only used union by rank? 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 merge two sets by size

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

C. To initialize the parent array

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 B

4. Quick Check:

   Path compression = flatten tree for faster find [OK]

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

Common Mistakes:

• Confusing path compression with union by size/rank
• Thinking it merges sets instead of optimizing find
• Assuming it initializes data structures

2. Which of the following is the correct Java method signature for the find function implementing path compression in Union Find?

easy

A. public int find(int x)

B. public void find(int x)

C. public boolean find(int x)

D. public int find()"

Solution

1. Step 1: Understand the return type of find

   The find method returns the representative (root) of the set containing element x, so it returns an integer.

2. Step 2: Check method parameters

   The method must accept the element x as input to find its root.

3. Final Answer:

   public int find(int x) -> Option A

4. Quick Check:

   find returns root index as int [OK]

Hint: find returns int root and takes int parameter [OK]

Common Mistakes:

• Using void return type for find
• Missing parameter for element to find
• Returning boolean instead of int

3. Consider the following Java code snippet implementing path compression in Union Find:

int[] parent = {0, 0, 1, 2, 3};

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

System.out.println(find(4));
System.out.println(Arrays.toString(parent));

What is the output?

medium

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

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

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

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

Solution

1. Step 1: Trace find(4) call with path compression

   parent[4]=3 -> find(3); parent[3]=2 -> find(2); parent[2]=1 -> find(1); parent[1]=0 -> find(0)=0 (base); parent[1]=0; parent[2]=0; parent[3]=0; parent[4]=0. Returns 0.

2. Step 2: Update parent array after path compression

   Entire path flattens: parent = [0, 0, 0, 0, 0].

3. Final Answer:

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

4. Quick Check:

   find(4) = 0, all path nodes point to root 0 [OK]

Hint: Path compression flattens entire path to root [OK]

Common Mistakes:

• Thinking only leaf updates (partial compression)
• Expecting no change in parent array
• Misreading recursion depth or base case

4. The following find method is intended to implement path compression but has a bug:

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

What is the bug and how to fix it?

medium

A. Missing path compression assignment; fix by adding parent[x] = find(parent[x]);

B. Incorrect base case; fix by changing parent[x] == x to parent[x] != x

C. Wrong return type; fix by changing return type to void

D. No bug; code is correct

Solution

1. Step 1: Identify missing path compression step

   The method recursively finds the root but does not update parent[x] to root, so no path compression happens.

2. Step 2: Fix by assigning compressed parent

   Adding parent[x] = find(parent[x]); before returning compresses the path.

3. Final Answer:

   Missing path compression assignment; fix by adding parent[x] = find(parent[x]); -> Option A

4. Quick Check:

   Assign parent[x] = find(parent[x]) for compression [OK]

Hint: Always assign parent[x] = find(parent[x]) to compress path [OK]

Common Mistakes:

• Forgetting to update parent[x]
• Changing base case incorrectly
• Returning void instead of int

5. Given a Union Find structure with parent = {0, 0, 1, 2, 4, 5} and rank = {0, 1, 2, 3, 4, 5}, after performing union(4, 5) and then find(5) with path compression, what will be the updated parent array?

hard

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

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

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

D. [0, 0, 1, 2, 5, 5]

Solution

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

   Ranks: rank[4]=4, rank[5]=5. Since rank[5] > rank[4], parent[4] = 5. But ranks given are increasing, so actually rank[5] is highest, so parent[4] = 5.

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

   find(5) returns 5 (root). Path compression updates parent of 5 to itself (no change). But parent[4] points to 5, so parent array is [0,0,1,2,5,5].

3. Step 3: Check options

   [0, 0, 1, 2, 4, 4] shows parent[5] = 4, which contradicts union by rank. [0, 0, 1, 2, 3, 4] no change. [0, 0, 1, 2, 3, 3] parent[5]=3 incorrect. [0, 0, 1, 2, 5, 5] parent[4]=5 and parent[5]=5 matches expected.

4. Final Answer:

   [0, 0, 1, 2, 5, 5] -> Option D

5. Quick Check:

   Union by rank sets parent of lower rank to higher [OK]

Hint: Union by rank attaches lower rank root to higher rank root [OK]

Common Mistakes:

• Confusing which parent to update in union
• Ignoring rank comparison
• Assuming parent array unchanged after union