Union Find Disjoint Set Data Structure in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

It helps to keep track of elements divided into groups called sets, and quickly find which set an element belongs to or merge two sets.

beginner

Explain the 'find' operation in Union Find Disjoint Set.

The 'find' operation returns the representative or leader of the set that an element belongs to. It helps check if two elements are in the same set.

beginner

What does the 'union' operation do in the Union Find Disjoint Set?

The 'union' operation merges two different sets into one by connecting their leaders, so elements from both sets become part of the same group.

intermediate

Why is path compression important in the Union Find Disjoint Set?

Path compression makes the 'find' operation faster by making every node on the path point directly to the leader, reducing the tree height.

intermediate

What is the role of 'union by rank' or 'union by size' in Union Find?

It helps keep the tree shallow by always attaching the smaller tree under the root of the larger tree, improving efficiency of operations.

What does the 'find' operation return in a Union Find Disjoint Set?

A The leader or representative of the set

B The size of the set

C The number of sets

D The parent of the element

The 'find' operation returns the leader or representative of the set that the element belongs to.

What is the main goal of the 'union' operation?

A To delete an element from a set

B To split a set into two

C To find the size of a set

D To merge two sets into one

The 'union' operation merges two different sets into a single set.

Which technique helps speed up the 'find' operation by flattening the tree?

A Breadth-first search

B Union by size

C Path compression

D Depth-first search

Path compression flattens the tree by making nodes point directly to the leader, speeding up future 'find' operations.

What does 'union by rank' help to achieve?

A Keeps the tree balanced by attaching smaller tree under larger

B Deletes the smaller tree

C Finds the rank of an element

D Splits the set into ranks

'Union by rank' attaches the smaller tree under the larger tree to keep the tree shallow and efficient.

If two elements have the same leader, what does it mean?

A They belong to different sets

B They belong to the same set

C They are not connected

D They are the leaders themselves

If two elements share the same leader, they belong to the same set.

Describe how the Union Find Disjoint Set data structure works and why it is useful.

Think about grouping elements and checking if they belong together.

Explain the optimizations 'path compression' and 'union by rank' in Union Find and how they improve performance.

Consider how trees can become tall and how to keep them short.

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