Path Compression in Union Find in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

int find(int x) {
    if (parent[x] != x) {
        parent[x] = [1];
    }
    return parent[x];
}

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

Afind(parent[x])

Bx

Cparent[x]

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting parent[x] to x directly without recursion.

Not applying path compression by skipping assignment.

Using parent[x] without recursion.

The find function recursively finds the root parent of x and applies path compression by setting parent[x] to the root.

2fill in blank

medium

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

DSA Java

void union(int a, int b) {
    int rootA = find(a);
    int rootB = find(b);
    if (rootA != rootB) {
        parent[[1]] = rootB;
    }
}

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

ArootB

Bb

Ca

DrootA

Attempts:

3 left

💡 Hint

Common Mistakes

Using a or b directly instead of their roots.

Setting parent of rootB to rootA instead of rootA to rootB.

Not checking if roots are different before union.

We set the parent of rootA to rootB to merge the two sets.

3fill in blank

hard

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

DSA Java

int find(int x) {
    if (parent[x] == x) {
        return x;
    }
    return [1];
}

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

Aparent[x] = find(parent[x])

Bx

Cparent[x]

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning parent[x] directly without recursion.

Returning x always without recursion.

Returning an incorrect expression like x + 1.

The function must recursively call find on parent[x] to find the root and apply path compression.

4fill in blank

hard

Fill both blanks to implement path compression in the find function.

DSA Java

int find(int x) {
    if (parent[x] != x) {
        parent[x] = [1];
    }
    return [2];
}

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

Afind(parent[x])

Bparent[x]

Cx

Dfind(x)

Attempts:

3 left

💡 Hint

Common Mistakes

Returning x instead of parent[x].

Assigning parent[x] to x instead of recursive find.

Using find(x) causing infinite recursion.

We assign parent[x] to the root found by find(parent[x]) and then return parent[x].

5fill in blank

hard

Fill all three blanks to implement union by rank with path compression.

DSA Java

void union(int a, int b) {
    int rootA = find(a);
    int rootB = find(b);
    if (rootA != rootB) {
        if (rank[rootA] < rank[[1]]) {
            parent[rootA] = [2];
        } else if (rank[rootA] > rank[rootB]) {
            parent[[3]] = rootA;
        } else {
            parent[rootB] = rootA;
            rank[rootA]++;
        }
    }
}

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

ArootB

DrootA

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing rootA and rootB in assignments.

Not updating rank when ranks are equal.

Assigning parent incorrectly causing cycles.

We compare ranks of rootA and rootB. If rank[rootA] < rank[rootB], parent[rootA] = rootB. If rank[rootA] > rank[rootB], parent[rootB] = rootA. Otherwise, parent[rootB] = rootA and increment rank[rootA].

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