Union Find Disjoint Set Data Structure in DSA C - Build from Scratch

Union Find Disjoint Set Data Structure

📖 Scenario: Imagine you have a group of friends, and you want to know if two friends are in the same friend circle. A friend circle means everyone is connected directly or through other friends. We will use a Union Find Disjoint Set data structure to keep track of these friend circles.

🎯 Goal: You will build a simple Union Find Disjoint Set in C that can join two friend circles and check if two friends belong to the same circle.

📋 What You'll Learn

Create an array called parent to represent each friend's parent in the set

Create an array called rank to keep track of the tree height for optimization

Write a function find that returns the root parent of a friend with path compression

Write a function union_sets that joins two friend circles using rank to keep trees short

Write code to check if two friends are in the same friend circle using find

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding Union Find helps in solving problems related to grouping, connectivity, and efficient merging in software engineering and competitive programming.

1

Create the parent and rank arrays

Create two arrays called parent and rank of size 5. Initialize parent so each friend is their own parent (0 to 4), and initialize rank with all zeros.

DSA C

# Create arrays parent and rank with initial values
# Your code here

Hint

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

2

Write the find function with path compression

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

DSA C

#include 

int parent[5];
int rank[5];

int find(int x) {
    // Write find function with path compression
    // Your code here
}

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

Hint

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

3

Write the union_sets function to join two sets

Write a function called union_sets that takes two integers a and b. Find their root parents using find. If they are different, attach the smaller rank tree under the bigger rank tree. If ranks are equal, attach b under a and increase rank[a] by 1.

DSA C

#include 

int parent[5];
int rank[5];

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

void union_sets(int a, int b) {
    // Write union_sets function to join sets
    // Your code here
}

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

Hint

Use find on both a and b. Compare ranks and attach the smaller tree under the bigger one. If equal, attach b under a and increase rank[a].

4

Check if two friends are in the same set and print result

In main, use union_sets to join friends 0 and 1, and friends 1 and 2. Then check if friends 0 and 2 are in the same set by comparing find(0) and find(2). Print "Yes" if they are in the same set, otherwise print "No".

DSA C

#include 

int parent[5];
int rank[5];

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

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

int main() {
    for (int i = 0; i < 5; i++) {
        parent[i] = i;
        rank[i] = 0;
    }
    // Use union_sets to join friends 0 and 1, and 1 and 2
    // Check if 0 and 2 are in the same set and print "Yes" or "No"
    // Your code here
    return 0;
}

Hint

Call union_sets(0, 1) and union_sets(1, 2). Then compare find(0) and find(2). Print "Yes" if equal, else "No".