Path Compression in Union Find in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the root parent of a node in Union Find.

DSA C++

int find(int x) {
    if (parent[x] == x) {
        return x;
    }
    return [1];
}

Drag options to blanks, or click blank then click option'

Aparent[x]

Bfind(parent[x])

Cx

Dx + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning parent[x] directly without recursion.

Returning x without checking if it's the root.

The find function recursively finds the root parent of x by calling find on its parent until it reaches the root.

2fill in blank

medium

Complete the code to apply path compression in the find function.

DSA C++

int find(int x) {
    if (parent[x] != x) {
        parent[x] = [1];
    }
    return parent[x];
}

Drag options to blanks, or click blank then click option'

Ax

Bfind(x)

Cfind(parent[x])

Dparent[parent[x]]

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to x itself.

Assigning parent[x] to find(x) causing infinite recursion.

Path compression sets the parent of x directly to the root parent found by recursively calling find on parent[x].

3fill in blank

hard

Fix the error in the union function to correctly merge two sets.

DSA C++

void unionSets(int a, int b) {
    a = find(a);
    b = find(b);
    if (a != b) {
        parent[a] = [1];
    }
}

Drag options to blanks, or click blank then click option'

Ab

Ba

Cfind(a)

Dfind(b)

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[a] to a itself, causing no change.

Assigning parent[a] to find(a), which is redundant.

To merge two sets, we set the parent of root a to root b, effectively connecting the two sets.

4fill in blank

hard

Fill both blanks to implement union by rank optimization.

DSA C++

void unionSets(int a, int b) {
    a = find(a);
    b = find(b);
    if (a != b) {
        if (rank[a] < rank[b]) {
            parent[a] = [1];
        } else if (rank[a] > rank[b]) {
            parent[b] = [2];
        } else {
            parent[b] = a;
            rank[a]++;
        }
    }
}

Drag options to blanks, or click blank then click option'

Ab

Ba

Crank[a]

Drank[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the parents incorrectly.

Not updating ranks properly.

If rank[a] is less, attach a to b. If rank[b] is less, attach b to a. This keeps trees shallow.

5fill in blank

hard

Fill all three blanks to implement path compression and union by rank in union-find.

DSA C++

int find(int x) {
    if (parent[x] != x) {
        parent[x] = [1];
    }
    return parent[x];
}

void unionSets(int a, int b) {
    a = find(a);
    b = find(b);
    if (a != b) {
        if (rank[a] < rank[b]) {
            parent[a] = [2];
        } else if (rank[a] > rank[b]) {
            parent[b] = [3];
        } else {
            parent[b] = a;
            rank[a]++;
        }
    }
}

Drag options to blanks, or click blank then click option'

Afind(parent[x])

Bb

Ca

Dfind(x)

Attempts:

3 left

💡 Hint

Common Mistakes

Not compressing paths in find.

Incorrectly assigning parents in union.

The find function uses path compression by setting parent[x] to the root found by find(parent[x]). The union function attaches the smaller rank tree under the bigger rank tree.

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