Path Compression in Union Find in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the root parent of a node with path compression.

DSA Python

def find(parent, x):
    if parent[x] != x:
        parent[x] = [1]  # Path compression step
    return parent[x]

Drag options to blanks, or click blank then click option'

Afind(parent, parent[x])

Bparent[x]

Cx

Dparent[parent[x]]

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to itself without recursion.

Using parent[parent[x]] which may not be the root.

Returning x directly without checking parent.

The path compression step recursively finds the root parent and updates the parent of x to point directly to the root.

2fill in blank

medium

Complete the union function to merge two sets by their root parents.

DSA Python

def union(parent, x, y):
    rootX = find(parent, x)
    rootY = find(parent, y)
    if rootX != rootY:
        parent[[1]] = rootY

Drag options to blanks, or click blank then click option'

ArootX

Bx

CrootY

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent of x or y directly instead of their roots.

Assigning parent of rootY to rootX instead of rootX to rootY.

We set the parent of rootX to rootY to merge the two sets.

3fill in blank

hard

Fix the error in the find function to correctly implement path compression.

DSA Python

def find(parent, x):
    if parent[x] != x:
        [1]  # Fix this line
    return parent[x]

Drag options to blanks, or click blank then click option'

Aparent[x] = find(parent, x)

Bparent[x] = find(parent, parent[x])

Cparent[x] = x

Dreturn x

Attempts:

3 left

💡 Hint

Common Mistakes

Calling find on x instead of parent[x].

Assigning parent[x] to x directly.

Returning x instead of updating parent[x].

The correct path compression updates parent[x] to the root found by recursively calling find on parent[x].

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each node to its root parent.

DSA Python

root_map = {node: [1] for node in nodes if [2]

Drag options to blanks, or click blank then click option'

Afind(parent, node)

Bnode in parent

Cnode not in parent

Dparent[node]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parent[node] directly without find.

Filtering nodes not in parent instead of those in parent.

We map each node to its root parent using find, and include only nodes present in parent.

5fill in blank

hard

Fill all three blanks to create a union-find initialization with path compression and union operations.

DSA Python

parent = {node: [1] for node in nodes}

def find(parent, x):
    if parent[x] != x:
        parent[x] = [2]
    return parent[x]

def union(parent, x, y):
    rootX = find(parent, x)
    rootY = find(parent, y)
    if rootX != rootY:
        parent[[3]] = rootY

Drag options to blanks, or click blank then click option'

Anode

Bfind(parent, parent[x])

CrootX

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing parent with wrong values.

Incorrect path compression assignment.

Assigning parent of x or y instead of rootX in union.

Initialize each node as its own parent, compress paths by updating parent[x] to the root, and union by setting parent of rootX to rootY.

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