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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of find operation with path compression

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

DSA Java

class UnionFind {
    int[] parent;
    public UnionFind(int n) {
        parent = new int[n];
        for (int i = 0; i < n; i++) parent[i] = i;
    }
    public int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }
    public void union(int a, int b) {
        int rootA = find(a);
        int rootB = find(b);
        if (rootA != rootB) parent[rootB] = rootA;
    }
}

public class Main {
    public static void main(String[] args) {
        UnionFind uf = new UnionFind(5);
        uf.union(0, 1);
        uf.union(1, 2);
        System.out.println(uf.find(2));
        System.out.println(java.util.Arrays.toString(uf.parent));
    }
}

A

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

B

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

C

0
[0, 0, 0, 3, 4]

D

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

💡 Hint

Remember that path compression updates the parent of all nodes on the find path to the root.

After union(0,1) and union(1,2), the root of 0,1,2 is 0. Calling find(2) compresses the path, setting parent[2] = 0. So parent array becomes [0,0,0,3,4].

🧠 Conceptual

intermediate

Effect of path compression on parent array

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

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

B [0, 1, 1, 3, 4]

C [0, 0, 2, 3, 4]

D [0, 0, 0, 3, 4]

💡 Hint

Path compression sets the parent of the node directly to the root.

Node 2's parent is 1, whose parent is 0. After find(2), parent[2] is updated to 0, so the array becomes [0, 0, 0, 3, 4].

🔧 Debug

advanced

Identify the bug in path compression implementation

What error will occur when running this Union Find find method with path compression?

DSA Java

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

A No error, works correctly

B StackOverflowError due to infinite recursion

C Compilation error: missing return statement

D NullPointerException at runtime

💡 Hint

Check if all code paths return a value.

The original method did not return a value after the recursive call, causing a compilation error for missing return statement. Adding 'return parent[x];' fixes the method.

❓ Predict Output

advanced

Output after multiple unions and finds with path compression

What is the output of the parent array after these operations?

DSA Java

UnionFind uf = new UnionFind(6);
uf.union(0, 1);
uf.union(1, 2);
uf.union(3, 4);
uf.union(4, 5);
uf.union(2, 5);
uf.find(5);
System.out.println(java.util.Arrays.toString(uf.parent));

A [0, 1, 2, 3, 4, 5]

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

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

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

💡 Hint

Trace unions carefully and remember path compression updates parents on find.

All nodes 0,1,2,3,4,5 become connected with root 0 after union(2,5). find(5) compresses path, setting parent of 3,4,5 to 0.

🧠 Conceptual

expert

Time complexity impact of path compression in Union Find

What is the amortized time complexity of the find operation with path compression in a Union Find data structure?

A O(α(n)) amortized, where α is the inverse Ackermann function

B O(1) amortized

C O(n) worst case

D O(log n) amortized

💡 Hint

Path compression combined with union by rank gives almost constant time.

The find operation with path compression has amortized time complexity O(α(n)), where α(n) is the inverse Ackermann function, which grows extremely slowly and is practically constant.

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