Path Compression in Union Find in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of path compression in Union Find?

Path compression helps to speed up the 'find' operation by making nodes on the path point directly to the root, flattening the structure for faster future queries.

beginner

In Union Find, what does the 'find' function do?

The 'find' function returns the representative or root of the set that a particular element belongs to.

intermediate

How does path compression affect the time complexity of Union Find operations?

Path compression reduces the time complexity of 'find' and 'union' operations to nearly O(1) amortized, specifically O(α(n)) where α is the inverse Ackermann function, which grows very slowly.

intermediate

Show the key line of code that performs path compression in the 'find' function.

return parent[node] = find(parent[node]); // This line sets the parent of 'node' directly to the root.

advanced

Why is path compression considered a 'lazy' optimization?

Because it only updates the parent pointers during 'find' calls, not immediately during union operations, making the structure flatter over time as needed.

What does path compression do in the Union Find data structure?

A Flattens the tree by making nodes point directly to the root during find operations

B Merges two sets by size

C Deletes nodes from the set

D Creates new sets for each element

Path compression flattens the tree by updating parent pointers during find, speeding up future queries.

Which function in Union Find is directly improved by path compression?

A Union

B Find

C Insert

D Delete

Path compression optimizes the 'find' function by flattening the tree structure.

What is the typical time complexity of Union Find operations with path compression?

A O(n)

B O(log n)

C O(n^2)

D O(α(n)) (inverse Ackermann function)

With path compression, Union Find operations run in almost constant time, O(α(n)), where α is the inverse Ackermann function.

In the path compression line 'parent[node] = find(parent[node]);', what does this assignment do?

A Sets node's parent to the root of its set

B Removes node from the set

C Sets node's parent to itself

D Creates a new set for node

This line updates the node's parent to the root, flattening the path.

Why is path compression done only during 'find' and not during 'union'?

A Because 'union' does not access parents

B Because 'union' is slower

C To keep union operation simple and defer flattening until needed

D Because path compression is unrelated to union

Path compression is a lazy optimization done during 'find' to keep 'union' simple and efficient.

Explain how path compression works in the Union Find data structure and why it improves performance.

Think about what happens to the parent pointers when you call find repeatedly.

Describe the difference between the 'union' operation and the 'find' operation in Union Find, especially focusing on where path compression is applied.

Consider which operation changes the structure and which one optimizes the search.

Practice

1. What is the main purpose of path compression in the Union Find data structure?

easy

A. To merge two sets by size

B. To speed up the find operation by making nodes point directly to the root

C. To initialize the parent array

D. To count the number of connected components

Solution

1. Step 1: Understand the find operation

   The find operation locates the root parent of a node in Union Find.

2. Step 2: Role of path compression

   Path compression updates nodes along the path to point directly to the root, reducing future search time.

3. Final Answer:

   To speed up the find operation by making nodes point directly to the root -> Option B

4. Quick Check:

   Path compression optimizes find = To speed up the find operation by making nodes point directly to the root [OK]

Hint: Path compression flattens trees to speed up find [OK]

Common Mistakes:

• Confusing path compression with union by size/rank
• Thinking it merges sets instead of optimizing find
• Assuming it initializes data structures

2. Which of the following is the correct syntax for the find function with path compression in C++?

easy

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

B. int find(int x) { while(parent[x] != x) x = parent[x]; return x; }

C. int find(int x) { return parent[x]; }

D. int find(int x) { if(parent[x] == x) return x; else return find(parent[x]); }

Solution

1. Step 1: Identify path compression usage

   Path compression updates parent[x] during recursion to root.

2. Step 2: Check each option

   int find(int x) { if(parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } updates parent[x] with recursive call result, implementing path compression correctly.

3. Final Answer:

   int find(int x) { if(parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } -> Option A

4. Quick Check:

   Path compression requires assignment in recursion = int find(int x) { if(parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } [OK]

Hint: Look for recursive assignment to parent[x] in find [OK]

Common Mistakes:

• Missing assignment to parent[x] for compression
• Using iterative find without compression
• Returning parent[x] without recursion

3. Consider the following code snippet using Union Find with path compression:

int parent[5] = {0, 1, 2, 3, 4};

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

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

unionSet(1, 2);
unionSet(2, 3);
find(3);

What is the state of the parent array after these operations?

medium

A. [0, 1, 3, 3, 4]

B. [0, 1, 2, 2, 4]

C. [0, 1, 1, 2, 4]

D. [0, 1, 1, 1, 4]

Solution

1. Step 1: Perform unions

   unionSet(1,2): parent[2] = 1; unionSet(2,3): parent[3] = find(2) = 1, so parent[3] = 1.

2. Step 2: Apply find(3) with path compression

   find(3) updates parent[3] to root 1 (already 1), so no change. parent array: [0,1,1,1,4]

3. Final Answer:

   [0, 1, 1, 1, 4] -> Option D

4. Quick Check:

   Parents after unions and compression = [0,1,1,1,4] [OK]

Hint: Trace unions first, then apply find with compression [OK]

Common Mistakes:

• Not updating parent of 3 during union
• Ignoring path compression effect in find
• Confusing indices and values in parent array

4. The following find function is intended to implement path compression but has a bug:

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

What is the bug in this code?

medium

A. It does not update parent[x] during recursion for compression

B. It uses == instead of != in the base case

C. It returns parent[x] instead of x

D. It causes infinite recursion

Solution

1. Step 1: Analyze recursion call

   The recursive call find(parent[x]) is made but its result is not assigned back to parent[x].

2. Step 2: Effect on path compression

   Without assignment, parent[x] remains unchanged, so path compression does not happen.

3. Final Answer:

   It does not update parent[x] during recursion for compression -> Option A

4. Quick Check:

   Missing assignment breaks path compression = It does not update parent[x] during recursion for compression [OK]

Hint: Always assign recursion result to parent[x] for compression [OK]

Common Mistakes:

• Thinking base case condition is wrong
• Assuming return value is incorrect
• Believing it causes infinite recursion

5. You have a Union Find structure with 6 elements (0 to 5). Initially, each element is its own parent. You perform these operations:

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

After all operations, what is the parent array state assuming path compression is used in find?

hard

A. [0, 1, 0, 3, 4, 3]

B. [0, 0, 0, 3, 3, 3]

C. [0, 0, 0, 0, 3, 0]

D. [0, 0, 1, 3, 3, 0]

Solution

1. Step 1: Perform initial unions

   unionSet(0,1): parent[1]=0; unionSet(1,2): parent[2]=find(1)=0; unionSet(3,4): parent[4]=3; unionSet(4,5): parent[5]=find(4)=3;

2. Step 2: Apply finds with path compression

   find(2): compresses path for 2 to 0 (already 0); find(5): compresses path for 5 to 3 (already 3); unionSet(2,5): parent[3]=find(2)=0; find(5): compresses path for 5 to 0, so parent[5]=0;

3. Step 3: Final parent array

   parent[0]=0, parent[1]=0, parent[2]=0, parent[3]=0 (updated), parent[4]=3, parent[5]=0 (compressed)

4. Final Answer:

   [0, 0, 0, 0, 3, 0] -> Option C

5. Quick Check:

   After unions and compression, parent array = [0, 0, 0, 0, 3, 0] [OK]

Hint: Track unions then compress paths after finds [OK]

Common Mistakes:

• Not updating parent of 3 during unionSet(2,5)
• Ignoring path compression effect on 5 after last find
• Mixing parent indices and values