Path Compression in Union Find in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the root parent of a node with path compression.

DSA Javascript

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

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

Afind(parent, parent[x])

Bparent[x]

Cx

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to itself without recursion.

Returning parent[x] before updating it.

Using a wrong base case.

The function recursively finds the root parent and updates the parent of x to that root, which is path compression.

2fill in blank

medium

Complete the union function to merge two sets by their root parents.

DSA Javascript

function union(parent, x, y) {
  let rootX = find(parent, x);
  let rootY = find(parent, y);
  if (rootX !== rootY) {
    [1] = rootY;
  }
}

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

Aparent[x]

BrootX

Cparent[rootX]

DrootY

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] instead of parent[rootX].

Assigning rootX to rootY instead of parent[rootX].

Not checking if roots are different.

We set the parent of rootX to rootY to merge the two sets.

3fill in blank

hard

Fix the error in the find function to correctly implement path compression.

DSA Javascript

function find(parent, 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.

Calling find with x instead of parent[x].

Assigning parent[x] to itself.

We must recursively call find on parent[x] to get the root and assign it to parent[x] for path compression.

4fill in blank

hard

Fill both blanks to complete the union function with path compression.

DSA Javascript

function union(parent, x, y) {
  let rootX = [1];
  let rootY = [2];
  if (rootX !== rootY) {
    parent[rootX] = rootY;
  }
}

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

Afind(parent, x)

Bx

Cfind(parent, y)

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Using x and y directly without finding roots.

Assigning rootX or rootY incorrectly.

We find the root parents of x and y before union to ensure correct merging.

5fill in blank

hard

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

DSA Javascript

function union(parent, rank, x, y) {
  let rootX = [1];
  let rootY = [2];
  if (rootX !== rootY) {
    if (rank[rootX] < rank[rootY]) {
      parent[rootX] = rootY;
    } else if (rank[rootX] > rank[rootY]) {
      parent[rootY] = rootX;
    } else {
      parent[rootY] = rootX;
      rank[rootX][3]1;
    }
  }
}

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

Afind(parent, x)

Bfind(parent, y)

C+=

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Not using find to get root parents.

Using wrong operator to update rank.

Not updating rank when ranks are equal.

We find root parents first, then update parent and rank correctly to keep tree balanced.

Practice

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

easy

A. To speed up the find operation by flattening the tree structure

B. To merge two sets without checking their roots

C. To initialize the parent array with default values

D. To count the number of connected components

Solution

1. Step 1: Understand the role of path compression

   Path compression is used during the find operation to make future queries faster by making nodes point directly to the root.

2. Step 2: Identify the effect on the tree structure

   This flattens the tree, reducing the height and speeding up subsequent find calls.

3. Final Answer:

   To speed up the find operation by flattening the tree structure -> Option A

4. Quick Check:

   Path compression = flatten tree = faster find [OK]

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

Common Mistakes:

• Confusing path compression with union by rank
• Thinking it merges sets without root checks
• Believing it initializes data structures

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

easy

A. function find(x) { if (parent[x] === x) return x; else return find(parent[x]); }

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

C. function find(x) { while (parent[x] !== x) x = parent[x]; return x; }

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

Solution

1. Step 1: Identify path compression in the code

   function find(x) { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } updates parent[x] to the root during recursion, which is path compression.

2. Step 2: Compare other options

   function find(x) { while (parent[x] !== x) x = parent[x]; return x; } uses iteration but no path compression. function find(x) { if (parent[x] === x) return x; else return find(parent[x]); } uses recursion but no assignment to parent[x]. function find(x) { parent[x] = x; return x; } resets parent incorrectly.

3. Final Answer:

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

4. Quick Check:

   Path compression requires parent update = function find(x) { if (parent[x] !== x) parent[x] = find(parent[x]); return parent[x]; } [OK]

Hint: Look for parent update inside find for path compression [OK]

Common Mistakes:

• Missing parent update for path compression
• Using iteration without compression
• Resetting parent incorrectly

3. Given the following code snippet, what will be the output of console.log(parent) after calling find(3)?

let parent = [0, 0, 1, 2, 4];
function find(x) {
  if (parent[x] !== x) parent[x] = find(parent[x]);
  return parent[x];
}
find(3);
console.log(parent);

medium

A. [0, 0, 0, 0, 4]

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

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

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

Solution

1. Step 1: Trace the find(3) call

   find(3) checks parent[3] = 2, not 3, so calls find(2). find(2) checks parent[2] = 1, not 2, so calls find(1). find(1) checks parent[1] = 0, not 1, so calls find(0). find(0) returns 0 since parent[0] = 0.

2. Step 2: Apply path compression updates

   parent[1] = 0, parent[2] = 0, parent[3] = 0 after recursion. But initial array was [0,0,1,2,4]. After updates, parent becomes [0,0,0,0,4]. However, the code only updates parent[x] during recursion. The first call updates parent[3] = find(2) = 0, parent[2] = find(1) = 0, parent[1] = find(0) = 0. So final parent array is [0,0,0,0,4].

3. Final Answer:

   [0, 0, 0, 0, 4] -> Option A

4. Quick Check:

   Path compression sets parents to root = [0,0,0,0,4] [OK]

Hint: Follow recursion and update parents to root [OK]

Common Mistakes:

• Not updating all parents during recursion
• Confusing parent indices
• Assuming no change after find

4. Identify the bug in the following find function implementing path compression:

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

medium

A. Function should use iteration instead of recursion

B. Incorrect base condition for recursion

C. Parent array is not updated for path compression

D. Missing return statement in the else branch

Solution

1. Step 1: Check the recursive calls and returns

   The else branch calls find(parent[x]) but does not return its result, so the function returns undefined for non-root nodes.

2. Step 2: Confirm missing return causes bug

   Without returning the recursive call, the function fails to return the root, breaking the find operation.

3. Final Answer:

   Missing return statement in the else branch -> Option D

4. Quick Check:

   Recursive calls must return values [OK]

Hint: Always return recursive find calls [OK]

Common Mistakes:

• Forgetting to return recursive call result
• Misunderstanding base case
• Thinking iteration is required

5. You have a Union Find structure with parent = [0, 0, 2, 2, 4] and rank = [0, 1, 0, 0, 0]. After performing union(3, 4) and then find(3), what will be the updated parent array assuming path compression is applied in find?

hard

A. [0, 0, 2, 2, 4]

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

C. [0, 0, 4, 4, 4]

D. [0, 0, 2, 4, 4]

Solution

1. Step 1: Perform union(3, 4)

   Find roots: find(3) = 2, find(4) = 4. Since rank[2] = 0 and rank[4] = 0, attach 3's root (2) to 4. So parent[2] = 4.

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

   find(3) calls find(2), which calls find(4). find(4) returns 4. Path compression sets parent[2] = 4 and parent[3] = parent[2] = 4.

3. Final Answer:

   [0, 0, 4, 4, 4] -> Option C

4. Quick Check:

   Union attaches root, find compresses path [OK]

Hint: Union attaches root; find compresses path to root [OK]

Common Mistakes:

• Not updating parent after union
• Ignoring path compression updates
• Confusing rank usage