Path Compression in Union Find in DSA Go - 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.

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

Scenario Under Consideration

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

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

This code finds the root of x and compresses the path by pointing x directly to the root.

Identify Repeating Operations

Look for repeated steps in the find function.

• Primary operation: Recursive calls following parent links up the tree.
• How many times: At most once per level from x to root, but path compression shortens future calls.

How Execution Grows With Input

Initially, trees can be tall, so find may take several steps.

Input Size (n)	Approx. Operations per find
10	Up to 4-5 steps, quickly reduced by compression
100	Up to 10-15 steps initially, then fewer
1000	Up to 20-30 steps initially, then much fewer

Pattern observation: The steps grow very slowly and flatten quickly due to path compression.

Final Time Complexity

Time Complexity: O(α(n))

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

Common Mistake

[X] Wrong: "Find always takes linear time because it may traverse many parents."

[OK] Correct: Path compression flattens the tree, so repeated finds become very fast, making the average cost almost constant.

Interview Connect

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

Self-Check

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

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 making nodes point directly to the root

B. To merge two sets by size

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 flatten the tree structure by making nodes point directly to the root.

2. Step 2: Compare with other operations

   Merging sets by size is done in union by size/rank, not path compression. Initialization and counting components are unrelated to path compression.

3. Final Answer:

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

4. Quick Check:

   Path compression optimizes find = To speed up the find operation by making nodes point directly to the root [OK]

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

Common Mistakes:

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

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

easy

A. func find(parent []int, x int) int { for parent[x] != x { x = parent[x] } return x }

B. func find(parent []int, x int) int { parent[x] = x; return x }

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

D. func find(parent []int, x int) int { if parent[x] != x { parent[x] = find(parent, parent[x]) } return parent[x] }

Solution

1. Step 1: Identify path compression in the code

   func find(parent []int, x int) int { if parent[x] != x { parent[x] = find(parent, parent[x]) } return parent[x] } recursively updates parent[x] to the root, which is path compression.

2. Step 2: Check other options for correctness

   func find(parent []int, x int) int { for parent[x] != x { x = parent[x] } return x } uses a loop but no path compression. func find(parent []int, x int) int { if parent[x] == x { return x } else { return find(parent, x) } } calls find without updating parent[x]. func find(parent []int, x int) int { parent[x] = x; return x } wrongly resets parent[x] to itself always.

3. Final Answer:

   func find(parent []int, x int) int { if parent[x] != x { parent[x] = find(parent, parent[x]) } return parent[x] } -> Option D

4. Quick Check:

   Recursive update of parent[x] = find(...) = func find(parent []int, x int) int { if parent[x] != x { parent[x] = find(parent, parent[x]) } return parent[x] } [OK]

Hint: Look for recursive parent update for path compression [OK]

Common Mistakes:

• Missing parent update in recursion
• Using loop without compression
• Resetting parent incorrectly

3. What will be the output of the following Go code snippet using Union Find with path compression?

package main
import "fmt"

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

func main() {
    parent := []int{0, 1, 2, 3, 4}
    parent[1] = 0
    parent[2] = 1
    fmt.Println(find(parent, 2))
    fmt.Println(parent)
}

medium

A. 1
[0 0 1 3 4]

B. 0
[0 0 0 3 4]

C. 2
[0 1 2 3 4]

D. 0
[0 1 0 3 4]

Solution

1. Step 1: Trace the find call for element 2

   find(2) calls find(1), which calls find(0). Since parent[0] = 0, root is 0.

2. Step 2: Apply path compression updates

   parent[2] is updated to 0, parent[1] is updated to 0. Final parent array is [0 0 0 3 4].

3. Final Answer:

   0 [0 0 0 3 4] -> Option B

4. Quick Check:

   Root of 2 is 0 and parent array updated = 0 [0 0 0 3 4] [OK]

Hint: Follow recursion and update parents on return [OK]

Common Mistakes:

• Not updating parent array after find
• Returning intermediate parent instead of root
• Confusing indices in parent array

4. Identify the bug in the following Go code implementing find with path compression:

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

medium

A. The function should not update parent[x] during recursion

B. The base case should check parent[x] != x instead of parent[x] == x

C. The recursive call should be find(parent, parent[x]) instead of find(parent, x)

D. The function should return x always

Solution

1. Step 1: Analyze the recursive call argument

   The recursive call uses find(parent, x), which causes infinite recursion because x does not change.

2. Step 2: Correct the recursive call

   It should call find(parent, parent[x]) to move up the tree towards the root.

3. Final Answer:

   The recursive call should be find(parent, parent[x]) instead of find(parent, x) -> Option C

4. Quick Check:

   Recursive call must move up tree = The recursive call should be find(parent, parent[x]) instead of find(parent, x) [OK]

Hint: Recursive find must use parent[x], not x [OK]

Common Mistakes:

• Calling find with same x causing infinite recursion
• Wrong base case condition
• Not updating parent[x] for compression

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

hard

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

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

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

D. [0, 0, 1, 3, 4]

Solution

1. Step 1: Trace the find call for element 4

   find(4) calls find(3), find(3) calls find(2), find(2) calls find(1), find(1) calls find(0). Root is 0.

2. Step 2: Apply path compression updates

   During recursion unwind: parent[1] = 0 (already), parent[2] = 0, parent[3] = 0, parent[4] = 0. Final parent array is [0, 0, 0, 0, 0].

3. Final Answer:

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

4. Quick Check:

   All nodes on path updated to root 0 = [0, 0, 0, 0, 0] [OK]

Hint: Update only nodes on path to root during find [OK]

Common Mistakes:

• Only updating parent[4] to root
• Failing to compress the full path
• Confusing parent indices