Path Compression in Union Find in DSA C++ - Build from Scratch

Path Compression in Union Find

📖 Scenario: Imagine you have a group of friends and you want to quickly find out if two people are in the same friend circle. We use a data structure called Union Find to help with this. Path Compression is a technique that makes finding the leader of a friend circle faster by shortening the path to the leader.

🎯 Goal: You will build a simple Union Find structure with path compression to quickly find the leader of any friend circle.

📋 What You'll Learn

Create an array called parent to represent each person's leader.

Create a function called find that uses path compression to find the leader of a person.

Create a function called union_sets to join two friend circles.

Print the leader of each person after some unions.

💡 Why This Matters

🌍 Real World

Union Find with path compression is used in social networks to quickly find connected groups, in network connectivity, and in clustering problems.

💼 Career

Understanding Union Find and path compression is important for software engineers working on graph algorithms, network design, and optimization problems.

1

Create the parent array

Create an integer array called parent of size 5 where each element is initialized to its own index (0 to 4).

DSA C++

// Create the parent array with 5 elements where parent[i] = i
// Your code here

Hint

Use a for loop from 0 to 4 and set parent[i] = i.

2

Write the find function with path compression

Write a function called find that takes an integer x and returns the leader of x. Use path compression by setting parent[x] to the leader during the search.

DSA C++

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

int find(int x) {
    // Your code here
}

Hint

Check if parent[x] is not x. If so, recursively find the leader and update parent[x].

3

Write the union_sets function

Write a function called union_sets that takes two integers a and b. Find their leaders using find and set the leader of b's group to the leader of a's group.

DSA C++

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

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

void union_sets(int a, int b) {
    // Your code here
}

Hint

Find leaders of a and b. If different, set parent[rootB] to rootA.

4

Print leaders after unions

Call union_sets(0, 1) and union_sets(1, 2). Then print the leader of each person from 0 to 4 using find. Use std::cout to print each leader on a new line.

DSA C++

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

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

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

union_sets(0, 1);
union_sets(1, 2);

// Print leaders of 0 to 4
// Your code here

Hint

After unions, the leader of 0, 1, and 2 should be the same. Print each leader on its own line.

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