Path Compression in Union Find in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of find with path compression

What is the output of the following code snippet that uses path compression in Union Find?

DSA Python

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))

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

uf = UnionFind(5)
uf.parent = [0, 0, 1, 2, 3]
print(uf.find(4))
print(uf.parent)

A

0
[0, 0, 0, 0, 0]

B

4
[0, 0, 0, 0, 0]

C

0
[0, 0, 1, 2, 3]

D

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

💡 Hint

Remember that path compression updates the parent of each node visited to the root.

The find function recursively finds the root of the element and updates the parent of each node along the path to point directly to the root. Initially, uf.parent is [0, 0, 1, 2, 3]. Calling find(4) traces the path 4 -> 3 -> 2 -> 1 -> 0. After path compression, all these nodes point directly to 0, so uf.parent becomes [0, 0, 0, 0, 0]. The root returned is 0.

🧠 Conceptual

intermediate

Effect of path compression on parent array

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

A [0, 0, 0, 2, 3]

B [0, 0, 1, 0, 3]

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

D [0, 0, 0, 0, 3]

💡 Hint

Path compression updates parents along the path to the root.

Calling find(3) follows the path 3 -> 2 -> 1 -> 0. After path compression, nodes 3, 2, and 1 will point directly to 0. So the parent array updates to [0, 0, 0, 0, 3].

🔧 Debug

advanced

Identify the bug in path compression implementation

What error will the following code produce when calling uf.find(3)?

DSA Python

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))

    def find(self, x):
        if self.parent[x] == x:
            return x
        self.parent[x] = self.find(self.parent[x])

uf = UnionFind(5)
uf.parent = [0, 0, 1, 2, 3]
print(uf.find(3))

A 0

B RecursionError

C None

D TypeError

💡 Hint

Check the return statements in the find method.

The find method is missing a return statement after the recursive call. So when self.parent[x] != x, the function returns None implicitly. Thus, uf.find(3) returns None.

❓ Predict Output

advanced

Union and find operations with path compression

What is the parent array after performing the following operations with path compression?

DSA Python

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))

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

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

uf = UnionFind(6)
uf.union(0, 1)
uf.union(1, 2)
uf.union(3, 4)
uf.union(4, 5)
uf.find(5)
print(uf.parent)

A [0, 0, 0, 3, 3, 3]

B ]3 ,3 ,3 ,0 ,0 ,0[

C 0, 0, 0, 3, 3, 3]

D [0, 0, 0, 3, 3, 3

💡 Hint

Trace union and find calls carefully, noting path compression effects.

Unions connect 0-1-2 and 3-4-5. After uf.find(5), path compression updates parent of 5 to 3. The parent array is [0, 0, 0, 3, 3, 3].

🧠 Conceptual

expert

Time complexity impact of path compression

Which statement best describes the time complexity impact of path compression in Union Find operations?

A Path compression makes each find operation run in constant O(1) time always.

B Path compression reduces the amortized time per operation to nearly O(1), specifically inverse Ackermann function time.

C Path compression increases the time complexity because it updates many parents during find.

D Path compression has no effect on time complexity; it only changes the parent array structure.

💡 Hint

Think about amortized complexity and how path compression optimizes repeated finds.

Path compression optimizes the find operation so that over many operations, the average time per operation is almost constant, bounded by the inverse Ackermann function, which grows extremely slowly.

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