Union Find Disjoint Set Data Structure in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the parent array so each element is its own parent.

DSA Java

for (int i = 0; i < n; i++) {
    parent[i] = [1];
}

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

An

Bi

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting all parents to 0 instead of their own index.

Using n as parent which is out of bounds.

Each element starts as its own parent, so parent[i] = i.

2fill in blank

medium

Complete the code to find the root parent of element x 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'

Ax

Bparent[parent[x]]

Cfind(parent[x])

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to parent[parent[x]] which may not be root.

Returning x without updating parent[x].

Path compression sets parent[x] to the root parent found recursively.

3fill in blank

hard

Fix the error in the union method to correctly merge two sets by rank.

DSA Java

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

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

Ab

Ba

Crank[a]

Drank[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping parents incorrectly causing cycles.

Not updating rank when ranks are equal.

When rank[a] < rank[b], a's parent is b; when rank[a] > rank[b], b's parent is a.

4fill in blank

hard

Fill both blanks to check if two elements are in the same set.

DSA Java

boolean connected(int x, int y) {
    return find(x) [1] find(y);
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using != which checks if they are in different sets.

Using < or > which are not meaningful here.

Two elements are connected if their root parents are equal.

5fill in blank

hard

Fill all three blanks to implement union by size instead of rank.

DSA Java

void unionBySize(int a, int b) {
    a = find(a);
    b = find(b);
    if (a != b) {
        if (size[a] < size[b]) {
            parent[a] = [1];
            size[[2]] += size[a];
        } else {
            parent[b] = [3];
            size[[3]] += size[b];
        }
    }
}

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

Ab

Ba

Attempts:

3 left

💡 Hint

Common Mistakes

Updating size of wrong root after union.

Swapping parent assignments incorrectly.

Attach smaller size tree under bigger; update size of new parent accordingly.

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