Path Compression in Union Find in DSA Python - Build from Scratch

Path Compression in Union Find

📖 Scenario: Imagine you are managing a social network where people form friend groups. You want to quickly find out if two people are in the same friend group and efficiently merge groups when new friendships form.

🎯 Goal: Build a Union Find data structure with path compression to efficiently find and union friend groups.

📋 What You'll Learn

Create a list called parent to represent each person's parent in the friend group.

Create a function find that uses path compression to find the root parent of a person.

Create a function union to merge two friend groups.

Use the find and union functions to process friendship connections.

Print the final parent list showing the root parent of each person after unions.

💡 Why This Matters

🌍 Real World

Union Find with path compression is used in social networks, network connectivity, and clustering to quickly find connected components.

💼 Career

Understanding Union Find is important for software engineers working on graph algorithms, network design, and optimization problems.

1

Create the initial parent list

Create a list called parent with 5 elements where each element is its own index (0 to 4). This means each person is initially their own parent.

DSA Python

# Create the parent list where each person is their own parent
# Your code here

Hint

Each person starts as their own group leader, so the parent list is just their own indices.

2

Define the find function with path compression

Define a function called find that takes a parameter x. It should return the root parent of x. Use path compression by setting parent[x] to the root parent during the search.

DSA Python

parent = [0, 1, 2, 3, 4]

# Define the find function with path compression
# Your code here

Hint

Use recursion to find the root and update parent[x] to the root to flatten the tree.

3

Define the union function to merge groups

Define a function called union that takes two parameters a and b. It should find the root parents of a and b using the find function and set the parent of b's root to a's root to merge the groups.

DSA Python

parent = [0, 1, 2, 3, 4]

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]

# Define the union function to merge two groups
# Your code here

Hint

Find the roots of both elements and link one root to the other to merge sets.

4

Use union and find, then print the parent list

Use the union function to merge these pairs: (0, 1), (1, 2), and (3, 4). Then print the parent list after these unions.

DSA Python

parent = [0, 1, 2, 3, 4]

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]

def union(a, b):
    rootA = find(a)
    rootB = find(b)
    if rootA != rootB:
        parent[rootB] = rootA

# Use union on (0, 1), (1, 2), (3, 4) and print parent list
# Your code here

Hint

After unions, the parent list shows the root parent for each person. Use print(parent) to display it.

Practice

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

easy

A. To delete elements from the set efficiently

B. To merge two sets without checking their roots

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

D. To sort the elements in each set

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 C

4. Quick Check:

   Path compression = speed up find [OK]

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

Common Mistakes:

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

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

easy

A. def find(x): parent[x] = x return x

B. def find(x): while parent[x] != x: x = parent[x] return x

C. def find(x): return x if parent[x] == x else find(parent[x])

D. def find(x): if parent[x] != x: parent[x] = find(parent[x]) return parent[x]

Solution

1. Step 1: Check for path compression in the code

   def 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: Verify other options

   def find(x): while parent[x] != x: x = parent[x] return x uses iteration but no path compression update. def find(x): return x if parent[x] == x else find(parent[x]) lacks assignment to parent[x]. def find(x): parent[x] = x return x incorrectly resets parent[x] always.

3. Final Answer:

   def find(x): if parent[x] != x: parent[x] = find(parent[x]) return parent[x] -> Option D

4. Quick Check:

   Path compression requires assignment in find [OK]

Hint: Path compression assigns parent[x] during find [OK]

Common Mistakes:

• Missing assignment to parent[x]
• Using iteration without updating parents
• Resetting parent[x] incorrectly

3. Given the following code snippet with path compression, what will be the output of print(parent) after the operations?

parent = [0, 1, 2, 3, 4]
def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]
def union(a, b):
    rootA = find(a)
    rootB = find(b)
    if rootA != rootB:
        parent[rootB] = rootA

union(1, 2)
union(2, 3)
find(3)
print(parent)

medium

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

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

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

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

Solution

1. Step 1: Perform unions and track parents

   union(1,2): parent[2] = 1; union(2,3): parent[3] = find(2) = 1, so parent[3] = 1.

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

   find(3) updates parent[3] to root 1 (already 1), so no change. Final parent array: [0,1,1,1,4]

3. Final Answer:

   [0, 1, 1, 1, 4] -> Option A

4. Quick Check:

   Parents after unions and find = [0,1,1,1,4] [OK]

Hint: Trace unions then find with path compression [OK]

Common Mistakes:

• Not updating parent during find
• Confusing root assignments
• Ignoring path compression effect

4. Identify the bug in the following find function that is supposed to implement path compression:

def find(x):
    if parent[x] == x:
        return x
    else:
        find(parent[x])

medium

A. Incorrect base case condition

B. Missing return statement in the else branch

C. Parent array is not initialized

D. Using recursion instead of iteration

Solution

1. Step 1: Check the else branch

   The else branch calls find(parent[x]) but does not return its result, so the function returns None.

2. Step 2: Understand path compression requirement

   Without returning the recursive call result, path compression and correct root finding fail.

3. Final Answer:

   Missing return statement in the else branch -> Option B

4. Quick Check:

   Return recursive find result [OK]

Hint: Always return recursive call in find [OK]

Common Mistakes:

• Forgetting return in recursion
• Misunderstanding base case
• Assuming iteration is required

5. You have a Union Find structure with 5 elements (0 to 4). After performing union(0,1), union(1,2), and union(3,4), you call find(2) and find(4) with path compression. Which of the following statements is true about the parent array after these operations?

hard

A. Elements 0, 1, and 2 will have the same root, and 3 and 4 will have the same root, with path compression applied

B. All elements will have different roots because path compression does not merge sets

C. Elements 0 and 4 will share the same root after path compression

D. Path compression will cause all elements to point to element 0

Solution

1. Step 1: Analyze unions

   union(0,1) and union(1,2) merge 0,1,2 into one set. union(3,4) merges 3 and 4 into another set.

2. Step 2: Effect of find with path compression

   find(2) compresses path so 2 points directly to root of set {0,1,2}. find(4) compresses path so 4 points directly to root of set {3,4}.

3. Step 3: Confirm roots and parent array

   Sets remain separate; 0,1,2 share root; 3,4 share root; path compression flattens trees.

4. Final Answer:

   Elements 0, 1, and 2 will have the same root, and 3 and 4 will have the same root, with path compression applied -> Option A

5. Quick Check:

   Unions merge sets; path compression flattens but does not merge sets [OK]

Hint: Unions merge sets; path compression flattens trees only [OK]

Common Mistakes:

• Thinking path compression merges sets
• Assuming all elements share one root
• Confusing union and find roles