#0242

Count Connected Components with Union-Find

MediumDSA28 min20 XP

Problem

Given n nodes labeled 0 to n-1 and undirected edges, return the number of connected components.

Why This Matters

Union-find is the fastest mental model for dynamic connectivity questions where edges merge groups.

Function Signature

def count_components(n, edges):

Examples

Example 1

Inputn = 5, edges = [[0, 1], [1, 2], [3, 4]]

Output2

Nodes 0-1-2 form one component and 3-4 form another.

Constraints

Return the exact requested value.
Handle empty or small inputs when the prompt allows them.
Prefer the target DSA pattern over brute force when the input can grow.

CodePython

def count_components(n, edges):
    """
    Input example: n = 5, edges = [[0, 1], [1, 2], [3, 4]]
    Return: 2
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return 0

def count_components(n, edges):
    """
    Input example: n = 5, edges = [[0, 1], [1, 2], [3, 4]]
    Return: 2
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return 0

Visible browser tests run here when available.

Testcases2 visible / 3 hidden categories

Two components

Inputn = 5, edges = [[0, 1], [1, 2], [3, 4]]

Expected2

Three unions reduce five nodes to two components.

Duplicate edge

Inputn = 3, edges = [[0, 1], [1, 0]]

Expected2

The second edge does not reduce the count again.

Hidden Test Categories

no edgesfully connected graphself-loop edge