Union Find Disjoint Set Data Structure in DSA C++ - Time & Space Complexity

Time Complexity: Union Find Disjoint Set Data Structure

O(α(n))

Understanding Time Complexity

We want to understand how fast the Union Find data structure works when connecting or checking groups.

How does the time needed change as we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class UnionFind {
  std::vector<int> parent, rank;
public:
  UnionFind(int n) : parent(n), rank(n, 0) {
    for (int i = 0; i < n; i++) parent[i] = i;
  }
  int find(int x) {
    if (parent[x] != x) parent[x] = find(parent[x]);
    return parent[x];
  }
  void unite(int a, int b) {
    a = find(a); b = find(b);
    if (a != b) {
      if (rank[a] < rank[b]) parent[a] = b;
      else if (rank[b] < rank[a]) parent[b] = a;
      else {
        parent[b] = a;
        rank[a]++;
      }
    }
  }
};
    

This code manages groups of items, quickly joining two groups or checking if two items share a group.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The find function uses recursion to find the group's leader.
• How many times: Each find call may recurse up the tree, but path compression shortens future calls.
• Union operation: Calls find twice and then merges groups, which is a constant-time step.

How Execution Grows With Input

As the number of items grows, each find call takes slightly longer at first but quickly becomes very fast due to path compression.

Input Size (n)	Approx. Operations per find/unite
10	About 2-3 steps
100	About 4-5 steps
1000	About 5-6 steps

Pattern observation: The steps grow very slowly, almost like a flat line, thanks to path compression and rank balancing.

Final Time Complexity

Time Complexity: O(α(n))

This means each operation runs almost in constant time, where α(n) is a very slow-growing function that is practically constant for all reasonable input sizes.

Common Mistake

[X] Wrong: "Each find operation takes linear time because it might traverse many parents."

[OK] Correct: Path compression flattens the structure, so repeated finds become very fast, avoiding long traversals.

Interview Connect

Understanding this time complexity shows you how efficient group management can be, a skill useful in many coding problems and real-world applications.

Self-Check

"What if we removed path compression from the find function? How would the time complexity change?"

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