Path Compression in Union Find in DSA C++ - Time & Space Complexity

Time Complexity: Path Compression in Union Find

O(α(n))

Understanding Time Complexity

We want to understand how fast the "find" operation runs when we use path compression in Union Find.

Specifically, how does the time grow as the number of elements increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int find(int x, vector<int>& parent) {
    if (parent[x] != x) {
        parent[x] = find(parent[x], parent);  // Path compression
    }
    return parent[x];
}

This code finds the root parent of element x and compresses the path to speed up future finds.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to find until the root parent is found.
• How many times: At most, the depth of the tree before compression.

How Execution Grows With Input

Initially, trees can be tall, so find might take several steps. But path compression flattens trees over time.

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

Pattern observation: The number of steps grows very slowly, almost constant, even as n grows large.

Final Time Complexity

Time Complexity: O(α(n))

This means each find operation runs in almost constant time, where α(n) is the very slow-growing inverse Ackermann function.

Common Mistake

[X] Wrong: "Path compression makes find run in constant time always."

[OK] Correct: While path compression greatly speeds up find, the time depends on α(n), which grows so slowly it looks constant but is not exactly constant.

Interview Connect

Understanding path compression's time complexity shows you how smart tweaks make algorithms very efficient in practice.

Self-Check

What if we removed path compression? How would the time complexity of find change?

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