ADVANCED DATA STRUCTURES PROGRAM • LEVEL 9 — DISJOINT SETS
Count Dynamic Connected Components
Learn how to count dynamic connected components using a clear C program.
PROBLEM UNDERSTANDING
Input and expected output
Sample input
No input required
Sample output
Components = 2
COMPLETE C PROGRAM
Complete C implementation
#include <stdio.h>
int parent[10], rank_value[10], set_size[10];
void initialize(int count) { for (int i = 0; i < count; i++) { parent[i] = i; rank_value[i] = 0; set_size[i] = 1; } }
int find_root(int value) { return parent[value] == value ? value : (parent[value] = find_root(parent[value])); }
void unite_rank(int first, int second)
{
first = find_root(first); second = find_root(second);
if (first == second) return;
if (rank_value[first] < rank_value[second]) { int temp = first; first = second; second = temp; }
parent[second] = first; set_size[first] += set_size[second];
if (rank_value[first] == rank_value[second]) rank_value[first]++;
}
int main(void)
{
initialize(6);
int components = 6, edges[][2] = {{0, 1}, {1, 2}, {3, 4}, {2, 4}};
for (int edge = 0; edge < 4; edge++) {
int first = find_root(edges[edge][0]), second = find_root(edges[edge][1]);
if (first != second) { unite_rank(first, second); components--; }
}
printf("Components = %d\n", components);
return 0;
}CURRENT STEP
SELECTED LINE
EXPECTED OUTPUT FOR THE SAMPLE
Components = 2
Step 0 of 0
PROGRAM EXPLANATION
Algorithm and explanation
- Read the required input values.
- Decrease the component counter only when an edge joins two different roots.
- Display the computed result.
Decrease the component counter only when an edge joins two different roots.
EFFICIENCY
Time and space complexity
Time complexity
O((n + m) alpha(n))
Auxiliary space
O(n)
DEBUGGING CHECKLIST
Common mistakes
Check this
Use the correct format specifier for every variable.
Check this
Initialize variables before using their values.
Check this
Check braces, semicolons and input order carefully.
Try it yourself
Practice: Run the program with the sample input, predict its output, and then test one boundary case of your own.
