Union Find Disjoint Set Data Structure in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the parent of each element to itself in the Union Find structure.

DSA C++

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

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

Ai

B0

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting all parents to 0 or -1 instead of their own index.

Using the size of the set instead of the element index.

Each element is initially its own parent, so parent[i] = i; sets this up.

2fill in blank

medium

Complete the code to find the root parent of element x with path compression.

DSA C++

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[x]

Cfind(parent[x])

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to itself causing infinite recursion.

Not using recursion to find the root parent.

Path compression sets parent[x] to the root parent found by recursively calling find(parent[x]).

3fill in blank

hard

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

DSA C++

void unionSets(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] = a;
        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

Assigning parent[a] = a or parent[b] = b which does nothing.

Mixing up which parent to assign in the if conditions.

When rank[a] < rank[b], set parent[a] = b; to attach the smaller tree under the bigger one, and vice versa.

4fill in blank

hard

Fill both blanks to implement union by size instead of rank.

DSA C++

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

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

Ab

Ba

Csize[a]

Dsize[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Using rank variables instead of size.

Assigning parents incorrectly causing cycles.

When size[a] < size[b], attach a under b by parent[a] = b;. Otherwise, attach b under a by parent[b] = a;.

5fill in blank

hard

Fill all three blanks to create a map of each element to its root parent only if the root is greater than 0.

DSA C++

std::map<int, int> rootMap;
for (int i = 1; i <= n; i++) {
    if ([3] > 0) {
        rootMap[[1]] = [2];
    }
}

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

Ai

Bfind(i)

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping keys and values.

Using element instead of root parent in the condition.

Map keys are elements i, values are their root parents find(i), and the condition checks if root parent is > 0.

Practice

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

easy

A. To efficiently track and merge groups of connected elements

B. To sort elements in ascending order

C. To store key-value pairs for fast lookup

D. To implement a queue with priority

Solution

1. Step 1: Understand the role of Union Find

   The Union Find data structure is designed to keep track of which elements belong to which groups or sets, especially when these groups can merge.

2. Step 2: Compare with other data structures

   Sorting elements, storing key-value pairs, or implementing priority queues are different tasks handled by other data structures.

3. Final Answer:

   To efficiently track and merge groups of connected elements -> Option A

4. Quick Check:

   Union Find = Track and merge groups [OK]

Hint: Union Find = track connected groups fast [OK]

Common Mistakes:

• Confusing Union Find with sorting algorithms
• Thinking it stores key-value pairs like a map
• Assuming it manages priority like a heap

2. Which of the following is the correct way to declare a parent array for Union Find of size n in C++?

easy

A. int parent[n+1]; for(int i=1; i<=n; i++) parent[i] = i;

B. int parent[n]; for(int i=0; i

C. int parent[n]; for(int i=0; i<=n; i++) parent[i] = i;

D. int parent[n]; for(int i=1; i

Solution

1. Step 1: Initialize parent array correctly

   The parent array should have size n, and each element should initially be its own parent, so indices 0 to n-1 must be set.

2. Step 2: Check loop boundaries

   Loop must run from 0 to n-1 (i < n) to avoid out-of-bounds errors.

3. Final Answer:

   int parent[n]; for(int i=0; i -> Option B

4. Quick Check:

   Initialize parents from 0 to n-1 [OK]

Hint: Initialize parents from 0 to n-1 [OK]

Common Mistakes:

• Using <= n causing out-of-bounds error
• Starting loop from 1 when array is 0-indexed
• Declaring array size incorrectly

3. Consider the following C++ code using Union Find:

int parent[5];
for(int i=0; i<5; i++) parent[i] = i;

// Union function
void unionSet(int a, int b) {
  int rootA = findSet(a);
  int rootB = findSet(b);
  if(rootA != rootB) parent[rootB] = rootA;
}

// Find function
int findSet(int x) {
  if(parent[x] == x) return x;
  return parent[x] = findSet(parent[x]);
}

unionSet(0, 1);
unionSet(1, 2);
unionSet(3, 4);

cout << (findSet(2) == findSet(0)) << " " << (findSet(4) == findSet(0));

What is the output?

medium

A. 0 0

B. 0 1

C. 1 1

D. 1 0

Solution

1. Step 1: Perform unions and find roots

   unionSet(0,1) merges sets of 0 and 1; unionSet(1,2) merges sets of 1 and 2, so 0,1,2 are connected. unionSet(3,4) connects 3 and 4.

2. Step 2: Check if 2 and 0 share the same root, and if 4 and 0 share the same root

   findSet(2) == findSet(0) is true (1), findSet(4) == findSet(0) is false (0).

3. Final Answer:

   1 0 -> Option D

4. Quick Check:

   Connected(2,0) = 1, Connected(4,0) = 0 [OK]

Hint: Check connected sets after unions [OK]

Common Mistakes:

• Assuming 4 is connected to 0
• Forgetting path compression in findSet
• Mixing up union order

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

int parent[5];
for(int i=0; i<5; i++) parent[i] = i;

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

void unionSet(int a, int b) {
  int rootA = findSet(a);
  int rootB = findSet(b);
  if(rootA != rootB) parent[rootB] = rootA;
}

unionSet(0, 1);
unionSet(1, 2);
cout << (findSet(2) == findSet(0));

medium

A. Incorrect initialization of parent array

B. Union function does not check if roots are equal

C. Missing path compression in findSet function

D. Array size is too small for elements

Solution

1. Step 1: Analyze findSet function

   The findSet function recursively finds the root but does not update parent[x] to root, missing path compression optimization.

2. Step 2: Understand impact of missing path compression

   Without path compression, findSet is slower on deep trees but still correct. Other options are incorrect because initialization and union checks are correct, and array size matches elements.

3. Final Answer:

   Missing path compression in findSet function -> Option C

4. Quick Check:

   findSet lacks path compression [OK]

Hint: Check if findSet updates parent for path compression [OK]

Common Mistakes:

• Thinking union lacks root equality check
• Assuming parent array is wrongly initialized
• Confusing array size with element count

5. You have 6 elements (0 to 5) initially in separate sets. After these operations:

unionSet(0, 1);
unionSet(2, 3);
unionSet(4, 5);
unionSet(1, 2);
unionSet(0, 4);

How many distinct sets remain?

hard

A. 1

B. 2

C. 3

D. 4

Solution

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

   Initially: {0}, {1}, {2}, {3}, {4}, {5} unionSet(0,1) -> {0,1}, {2}, {3}, {4}, {5} unionSet(2,3) -> {0,1}, {2,3}, {4}, {5} unionSet(4,5) -> {0,1}, {2,3}, {4,5} unionSet(1,2) merges {0,1} and {2,3} -> {0,1,2,3}, {4,5} unionSet(0,4) merges {0,1,2,3} and {4,5} -> {0,1,2,3,4,5}

2. Step 2: Count remaining distinct sets

   All elements are connected in one set after last union.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   All merged into one set [OK]

Hint: Count merges to find final distinct sets [OK]

Common Mistakes:

• Counting sets before all unions
• Not merging transitive connections
• Assuming unions only connect pairs