Path Compression in Union Find in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of find operation with path compression

What is the output of the following code snippet that performs find operations with path compression on a Union Find structure?

DSA C

int parent[6] = {0, 0, 1, 2, 3, 4};

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

int main() {
    printf("%d\n", find(5));
    printf("%d\n", parent[5]);
    return 0;
}

A

5
5

B

4
4

C

0
0

D

1
0

💡 Hint

Trace the parent array and see how path compression updates parent[5].

The initial parent array is [0, 0, 1, 2, 3, 4]. The find(5) call recursively finds the root parent of 5:

• parent[5] = 4, so find(4) is called.
• parent[4] = 3, so find(3) is called.
• parent[3] = 2, so find(2) is called.
• parent[2] = 1, so find(1) is called.
• parent[1] = 0, so find(0) is called.
• parent[0] = 0, so root is 0.

Path compression sets parent of all nodes in the path directly to 0. So parent[5] becomes 0.

Output is 0 (root) and parent[5] after compression is 0.

🧠 Conceptual

intermediate

Effect of path compression on parent array

Given a Union Find parent array: [0, 0, 1, 2, 3, 4], after calling find(5) with path compression, what will be the updated parent array?

A [0, 0, 1, 2, 3, 0]

B [0, 0, 0, 0, 0, 0]

C [0, 1, 2, 3, 4, 5]

D [0, 0, 1, 1, 3, 4]

💡 Hint

Path compression updates parents along the path to the root.

Calling find(5) triggers recursive path compression, setting the parent of every node on the path 5→4→3→2→1→0 directly to the root 0. Thus, parent[1..5] all become 0 (parent[1] already was). Updated array: [0, 0, 0, 0, 0, 0].

🔧 Debug

advanced

Identify the bug in path compression implementation

What error will the following code cause when calling find(3)?

DSA C

int parent[4] = {0, 0, 1, 2};

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

int main() {
    printf("%d\n", find(3));
    return 0;
}

A Correct output: 0

B Runtime error: segmentation fault

C Compilation error: missing return statement

D Stack overflow due to infinite recursion

💡 Hint

Check if path compression is applied and if parent array changes.

The find function correctly computes the root (0) for node 3 via recursion: 3→2→1→0, stopping since parent[0] == 0. However, it lacks path compression as there is no parent[x] = find(parent[x]) update. No errors occur: compiles fine, runs without segfault or stack overflow, outputs 0.

The implementation bug is missing path compression, but the question asks for errors caused by calling find(3), of which there are none.

❓ Predict Output

advanced

Output after union and find with path compression

What is the output of the following code snippet?

DSA C

int parent[5] = {0, 1, 2, 3, 4};

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

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

int main() {
    union_set(0, 1);
    union_set(1, 2);
    union_set(3, 4);
    printf("%d\n", find(2));
    printf("%d\n", parent[2]);
    printf("%d\n", find(4));
    return 0;
}

A

0
0
3

B

1
1
3

C

0
1
4

D

2
0
4

💡 Hint

Trace unions and path compression updates carefully.

Unions:

• union_set(0,1): parent[1] = 0
• union_set(1,2): find(1) = 0, find(2) = 2, parent[2] = 0
• union_set(3,4): parent[4] = 3

Find(2): parent[2] = 0, so root is 0. Path compression sets parent[2] = 0.

parent[2] after find is 0.

Find(4): parent[4] = 3, root is 3.

🧠 Conceptual

expert

Time complexity impact of path compression

Which statement best describes the time complexity impact of path compression in Union Find operations?

A Path compression makes union operations slower but find operations faster.

B Path compression increases the worst-case time complexity to linear per find operation.

C Path compression has no effect on time complexity but reduces memory usage.

D Path compression reduces the amortized time per find operation to nearly constant, specifically inverse Ackermann function time.

💡 Hint

Consider how path compression flattens the tree structure over multiple operations.

Path compression flattens the tree, making future find operations faster. The amortized time per find becomes almost constant, bounded by the inverse Ackermann function, which grows extremely slowly.