Union Find Disjoint Set Data Structure in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Union Find after union operations

What is the output of the parent array after performing the union operations shown below on a Union Find structure of size 5?

DSA Java

int[] parent = {0, 1, 2, 3, 4};
int[] rank = {0, 0, 0, 0, 0};

// union(0, 1)
// union(1, 2)
// union(3, 4)

// After these unions, print the parent array.

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

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

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

D [1, 1, 1, 4, 4]

💡 Hint

Remember union by rank attaches the smaller tree under the root of the larger tree.

Initially, each element is its own parent.

union(0,1) makes 0 the parent of 1.

union(1,2) finds root of 1 is 0, so union(0,2) makes 0 the parent of 2.

union(3,4) makes 3 the parent of 4.

Final parent array: [0,0,0,3,3]

🧠 Conceptual

intermediate

Find operation result in Union Find

Given a Union Find structure with parent array [0, 0, 0, 3, 3], what is the result of find(2)?

A 0

B 3

C 2

D 1

💡 Hint

Follow the parent links until you reach the root.

find(2) checks parent[2] = 0, which is root since parent[0] = 0.

So, find(2) returns 0.

🔧 Debug

advanced

Identify the error in Union Find union method

What error will this union method cause when called with union(2, 3) on a Union Find of size 5?

DSA Java

void union(int x, int y) {
  int rootX = find(x);
  int rootY = find(y);
  if (rootX == rootY) return;
  if (rank[rootX] > rank[rootY]) {
    parent[rootY] = rootX;
  } else if (rank[rootX] < rank[rootY]) {
    parent[rootX] = rootY;
  } else {
    parent[rootX] = rootY;
  }
}

A No error, union works correctly

B Rank is not updated when ranks are equal, causing incorrect tree height

C NullPointerException due to uninitialized parent array

D StackOverflowError due to infinite recursion in find

💡 Hint

Check what happens when ranks are equal after union.

When ranks are equal, the code attaches rootX under rootY but does not increase rank[rootY].

This causes rank to be incorrect and may degrade performance.

📝 Syntax

advanced

Syntax error in Union Find find method

Which option contains the correct syntax for the find method with path compression?

A

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

B

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

C

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

D

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

💡 Hint

Check for missing parentheses and braces.

Option D uses correct Java syntax with braces and parentheses.

Option D misses parentheses in if condition causing syntax error.

Option D misses braces but is valid Java syntax (single statement if), but path compression requires braces to assign parent[x].

Option D returns without path compression update.

🚀 Application

expert

Number of connected components after unions

Given a Union Find of size 7 with these union operations: union(0,1), union(1,2), union(3,4), union(5,6), what is the number of connected components?

A 3

B 5

C 4

D 2

💡 Hint

Count how many groups are formed after unions.

Unions form groups: {0,1,2}, {3,4}, {5,6}.

Remaining single elements: { } (none).

Total connected components = 3.

Practice

1. What is the primary purpose of the Union Find Disjoint Set data structure?

easy

A. To efficiently manage and query connected components in a set

B. To sort elements in ascending order

C. To store key-value pairs for fast lookup

D. To implement a queue with fast enqueue and dequeue

Solution

1. Step 1: Understand the purpose of Union Find

   The Union Find data structure is designed to keep track of elements partitioned into disjoint sets and quickly find which set an element belongs to.

2. Step 2: Compare with other options

   Sorting, key-value storage, and queue operations are unrelated to Union Find's main goal.

3. Final Answer:

   To efficiently manage and query connected components in a set -> Option A

4. Quick Check:

   Union Find = Manage connected components [OK]

Hint: Union Find tracks groups, not order or key-value pairs [OK]

Common Mistakes:

• Confusing Union Find with sorting algorithms
• Thinking it stores key-value pairs like a map
• Assuming it implements queue operations

2. Which of the following is the correct way to initialize a Union Find structure for 5 elements in Java?

easy

A. UnionFind uf = new UnionFind[5];

B. UnionFind uf = new UnionFind(); uf.init(5);

C. UnionFind uf = new UnionFind(5);

D. UnionFind uf = new UnionFind(0, 5);

Solution

1. Step 1: Recognize typical Union Find constructor

   Usually, Union Find is initialized by passing the number of elements to its constructor, like new UnionFind(5).

2. Step 2: Check other options for syntax errors

   UnionFind uf = new UnionFind(); uf.init(5); suggests a separate init method which is uncommon; UnionFind uf = new UnionFind[5]; tries to create an array of UnionFind objects, not a single instance; UnionFind uf = new UnionFind(0, 5); uses two parameters which is not standard.

3. Final Answer:

   UnionFind uf = new UnionFind(5); -> Option C

4. Quick Check:

   Constructor with size parameter = UnionFind uf = new UnionFind(5); [OK]

Hint: Union Find needs size in constructor, no separate init [OK]

Common Mistakes:

• Trying to create an array of UnionFind instead of one instance
• Using multiple parameters in constructor incorrectly
• Assuming an init method is required after construction

3. Consider the following Java code using Union Find:

UnionFind uf = new UnionFind(5);
uf.union(0, 1);
uf.union(1, 2);
boolean connected = uf.connected(0, 2);
System.out.println(connected);

What will be the output?

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Trace union operations

   First, 0 and 1 are joined, then 1 and 2 are joined. This means 0, 1, and 2 belong to the same set.

2. Step 2: Check connected query

   Querying if 0 and 2 are connected returns true because they share the same root after unions.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Union chains connect 0 and 2 = true [OK]

Hint: Union chains connect elements; connected returns true [OK]

Common Mistakes:

• Assuming union does not link transitive elements
• Confusing union with overwrite of sets
• Expecting false because 0 and 2 were not directly united

4. The following code snippet has a bug. What is the error?

UnionFind uf = new UnionFind(3);
uf.union(0, 1);
uf.union(1, 2);
System.out.println(uf.connected(0, 3));

medium

A. Index out of bounds error at connected(0, 3)

B. No error, prints false

C. Compilation error due to wrong method call

D. Logical error, should print true

Solution

1. Step 1: Check the size and indices used

   The UnionFind is initialized with size 3, so valid indices are 0, 1, and 2. Index 3 is invalid.

2. Step 2: Identify the error at connected call

   Calling connected(0, 3) accesses an invalid index, causing an index out of bounds error.

3. Final Answer:

   Index out of bounds error at connected(0, 3) -> Option A

4. Quick Check:

   Invalid index access = Index out of bounds error [OK]

Hint: Check indices are within 0 to size-1 [OK]

Common Mistakes:

• Assuming connected returns false for invalid index
• Ignoring array bounds in Union Find
• Thinking compilation error occurs instead of runtime error

5. You have 6 elements (0 to 5). After performing unions: union(0,1), union(2,3), union(4,5), union(1,2), how many distinct sets remain?

hard

A. 3

B. 2

C. 1

D. 4

Solution

1. Step 1: Analyze unions step-by-step

   Initially, sets are {0}, {1}, {2}, {3}, {4}, {5}. - union(0,1) merges {0} and {1} -> {0,1} - union(2,3) merges {2} and {3} -> {2,3} - union(4,5) merges {4} and {5} -> {4,5} - union(1,2) merges sets containing 1 and 2 -> merges {0,1} and {2,3} -> {0,1,2,3}

2. Step 2: Count remaining distinct sets

   After all unions, sets are {0,1,2,3} and {4,5}, so total 2 sets remain.

3. Step 3: Re-examine options

   Options say 3, 2, 1, 4. The correct count is 2.

4. Final Answer:

   2 -> Option B

5. Quick Check:

   Union merges sets, final count = 2 [OK]

Hint: Count sets after unions by merging connected groups [OK]

Common Mistakes:

• Counting unions instead of distinct sets
• Forgetting transitive merges
• Assuming each union reduces sets by one without checking connections