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 array so each element is its own parent.

DSA C

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

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

An

B0

C-1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using n as parent which is out of array 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 C

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

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Afind(parent[x])

Bparent[parent[x]]

Cparent[x]

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

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

Returning x directly without path compression.

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

3fill in blank

hard

Fix the error in the union function to correctly merge two sets by their root parents.

DSA C

void union_sets(int a, int b) {
    a = find(a);
    b = find(b);
    if (a != b) {
        parent[[1]] = [2];
    }
}

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

Aa

Bb

Cparent[a]

Dparent[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[parent[a]] or parent[parent[b]] which is incorrect.

Assigning parent[b] = a instead of parent[a] = b.

We set the parent of root a to root b to merge sets.

4fill in blank

hard

Fill both blanks to implement union by rank optimization.

DSA C

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

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

Aa

Bb

Cparent[a]

Dparent[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping a and b incorrectly in assignment.

Not updating rank when ranks are equal.

If rank[a] < rank[b], set parent[a] = b to attach the smaller rank tree under the larger rank tree.

5fill in blank

hard

Fill all three blanks to implement a function that checks if two elements are in the same set.

DSA C

int is_same_set(int a, int b) {
    return [1](a) == [2](b);
}

void make_set(int v) {
    parent[v] = v;
    rank[v] = [3];
}

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

Afind

Bunion_sets

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using union_sets instead of find in is_same_set.

Initializing rank to 1 instead of 0.

Use find to get root parents and compare; rank is initialized to 0.

Practice

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

easy

A. To sort elements in ascending order

B. To efficiently track and merge groups of connected elements

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 merge.

2. Step 2: Compare with other data structures

   Sorting, key-value storage, and priority queues serve different purposes unrelated to grouping connected components.

3. Final Answer:

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

4. Quick Check:

   Union Find = track and merge groups [OK]

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

Common Mistakes:

• Confusing Union Find with sorting algorithms
• Thinking it stores key-value pairs
• Assuming it manages priority queues

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

easy

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

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

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

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

Solution

1. Step 1: Understand initialization of Union Find

   Each element starts as its own parent, so parent[i] = i for all i from 0 to n-1.

2. Step 2: Check each option

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

3. Final Answer:

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

4. Quick Check:

   Initialize parent[i] = i [OK]

Hint: Parent array starts with self indices [OK]

Common Mistakes:

• Setting all parents to zero
• Using -1 as parent without meaning
• Assigning parent to n (out of range)

3. Consider the following C code using Union Find for 5 elements (0 to 4):

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

// Union operations
union_set(parent, 0, 1);
union_set(parent, 1, 2);
union_set(parent, 3, 4);

// Find operations
int a = find_set(parent, 2);
int b = find_set(parent, 4);

printf("%d %d", a, b);

Assuming union_set merges sets and find_set returns the root parent, what is the printed output?

medium

A. 0 3

B. 1 3

C. 2 4

D. 0 4

Solution

1. Step 1: Trace union operations

   Initially, parent[i] = i for i=0..4.
   union_set(0,1) merges sets containing 0 and 1, root becomes 0.
   union_set(1,2) merges sets containing 1 and 2, root is 0 (since 1's root is 0).
   union_set(3,4) merges sets containing 3 and 4, root becomes 3.

2. Step 2: Trace find operations

   find_set(2) returns root of 2, which is 0.
   find_set(4) returns root of 4, which is 3.

3. Final Answer:

   0 3 -> Option A

4. Quick Check:

   Roots after unions: 2->0, 4->3 [OK]

Hint: Follow union roots to find final parents [OK]

Common Mistakes:

• Assuming find_set returns the element itself
• Not updating roots after unions
• Mixing up which element is root

4. The following C code attempts to implement union_set but has a bug:

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

What is the bug in this code?

medium

A. It should assign parent[rootB] = rootA instead

B. It does not check if rootA equals rootB before union

C. It assigns parent[a] instead of parent[rootA], breaking the set structure

D. It uses find_set incorrectly by passing wrong parameters

Solution

1. Step 1: Analyze union assignment

   The union should link the root of one set to the root of the other.
   Assigning parent[a] = rootB is wrong because 'a' may not be a root.

2. Step 2: Correct union assignment

   It should assign parent[rootA] = rootB to merge sets properly.
   This maintains the tree structure of the disjoint sets.

3. Final Answer:

   It assigns parent[a] instead of parent[rootA], breaking the set structure -> Option C

4. Quick Check:

   Union must link roots, not elements [OK]

Hint: Always link root parents in union [OK]

Common Mistakes:

• Assigning parent of non-root element
• Skipping root equality check
• Confusing parameters in find_set

5. You have 7 elements (0 to 6) initially separate. After these union operations:

union_set(parent, 0, 1);
union_set(parent, 1, 2);
union_set(parent, 3, 4);
union_set(parent, 5, 6);
union_set(parent, 2, 5);

How many distinct sets remain?

hard

A. 5

B. 4

C. 3

D. 2

Solution

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

   Initially: 7 sets: {0}, {1}, {2}, {3}, {4}, {5}, {6}
   union(0,1): {0,1}, {2}, {3}, {4}, {5}, {6}
   union(1,2): {0,1,2}, {3}, {4}, {5}, {6}
   union(3,4): {0,1,2}, {3,4}, {5}, {6}
   union(5,6): {0,1,2}, {3,4}, {5,6}
   union(2,5): merges sets {0,1,2} and {5,6} into {0,1,2,5,6}, {3,4}

2. Step 2: Count remaining sets

   Remaining sets are {0,1,2,5,6} and {3,4}.
   Total distinct sets = 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Count sets after unions = 2 [OK]

Hint: Count merged groups carefully after each union [OK]

Common Mistakes:

• Forgetting last union merges two sets
• Counting elements instead of sets
• Assuming all unions merge all sets