Path Compression in Union Find in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the parent of element x in the Union Find structure.

DSA C

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

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

Afind(parent, parent[x])

Bfind(parent, x)

Cx

Dparent[x]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning x directly without recursion.

Returning parent[x] without recursion.

The find function recursively finds the root parent of x by calling find on parent[x].

2fill in blank

medium

Complete the code to implement path compression by updating parent[x] during find.

DSA C

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

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

Ax

Bfind(parent, x)

Cparent[x]

Dfind(parent, parent[x])

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to x directly.

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

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

3fill in blank

hard

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

DSA C

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

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

Aa

Bb

Cparent[a]

Dparent[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[a] = a which does nothing.

Assigning parent[b] = b which does nothing.

To merge sets, assign parent of root b to root a: parent[b] = a. Fill the blank with 'a'.

4fill in blank

hard

Fill both blanks to complete the union function that merges two sets by their roots.

DSA C

void union_sets(int parent[], int a, int b) {
    a = find(parent, a);
    b = find(parent, b);
    if (a != b) {
        parent[[1]] = [2];
    }
}

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

Aa

Bb

Cparent[a]

Dparent[b]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order and assigning parent[a] = b.

Using parent[a] or parent[b] as indices.

To merge sets, assign parent of root b to root a: parent[b] = a.

5fill in blank

hard

Fill all three blanks to implement find with path compression and union that merges sets.

DSA C

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

void union_sets(int parent[], int a, int b) {
    a = [2];
    b = [3];
    if (a != b) {
        parent[b] = a;
    }
}

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

Afind(parent, parent[x])

Bfind(parent, a)

Cfind(parent, b)

Dfind(parent, x)

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating a and b in union.

Assigning parent[x] incorrectly in find.

In find, path compression assigns parent[x] to find(parent, parent[x]). In union, a and b are updated by find(parent, a) and find(parent, b) respectively.

Practice

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

easy

A. To initialize the parent array

B. To merge two sets by size

C. To count the number of connected components

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

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 makes all nodes on the path point directly to the root, speeding up future find calls.

3. Final Answer:

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

4. Quick Check:

   Path compression optimizes find = C [OK]

Hint: Path compression flattens tree for faster find [OK]

Common Mistakes:

• Confusing path compression with union by size
• Thinking it initializes the structure
• Believing it counts components

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

easy

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

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

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

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

Solution

1. Step 1: Identify path compression in code

   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 find 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 B

4. Quick Check:

   Path compression updates parent recursively = A [OK]

Hint: Look for parent update inside find recursion [OK]

Common Mistakes:

• Using loop without updating parent (Option B)
• Not updating parent during recursion (Option C)
• Resetting parent incorrectly (Option D)

3. Given the following Union Find structure with parent = [0, 0, 1, 2, 3], what will be the parent array after calling find(4) with path compression?

medium

A. [0, 0, 0, 0, 0]

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

C. [0, 1, 1, 2, 3]

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

Solution

1. Step 1: Trace find(4) path

   4's parent is 3, 3's parent is 2, 2's parent is 1, 1's parent is 0, which is root.

2. Step 2: Apply path compression

   All nodes on path (4,3,2,1) update parent to root 0 (parent[1] already 0). parent[2]=0, parent[3]=0, parent[4]=0. So parent array becomes [0,0,0,0,0].

3. Final Answer:

   [0, 0, 0, 0, 0] -> Option A

4. Quick Check:

   Path compression sets parent[4] = 0 = A [OK]

Hint: Update all parents on path to root after find [OK]

Common Mistakes:

• Only updating parent[4] to root (Option B)
• Not updating any parent (Option C)
• Updating wrong parents (Option D)

4. Identify the bug in this find function with path compression:

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

medium

A. No bug, code is correct

B. Incorrect base case condition

C. Missing return value from recursive call before assignment

D. Parent update should be after return

Solution

1. Step 1: Analyze recursive call

   The recursive call find(parent[x]) is made but its return value is ignored.

2. Step 2: Check parent update

   Parent update uses parent[parent[x]] instead of the root returned by recursion, so path compression fails.

3. Final Answer:

   Missing return value from recursive call before assignment -> Option C

4. Quick Check:

   Return value must be assigned to parent[x] = D [OK]

Hint: Assign recursive find result to parent[x] for compression [OK]

Common Mistakes:

• Ignoring recursive return value
• Updating parent incorrectly
• Misunderstanding base case

5. You have a Union Find with 5 elements and parent = [0, 0, 1, 2, 3]. After calling union(4, 0) and then find(4) with path compression, what is the final parent array?

hard

A. [0, 0, 1, 2, 0]

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

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

D. [0, 1, 1, 2, 0]

Solution

1. Step 1: Perform union(4, 0)

   Find roots: root(4) = 4 (chain 4->3->2->1->0), root(0) = 0. Since roots differ, link root(4) to root(0). So parent[4] = 0. Updated parent = [0, 0, 1, 2, 0].

2. Step 2: Call find(4) with path compression

   Path: 4 -> 0. Since 0 is root, parent[4] already 0, no further updates. Final parent = [0, 0, 1, 2, 0].

3. Final Answer:

   [0, 0, 1, 2, 0] -> Option A

4. Quick Check:

   Union links 4 to 0, find compresses path = A [OK]

Hint: Union links roots, find compresses path to root [OK]

Common Mistakes:

• Incorrectly linking assuming same root (Option C)
• Wrong union target (Option B)
• Incomplete path compression (Option D)