My solution for the roads and libraries graph problem. https://www.hackerrank.com/challenges/torque-and-development

roadsAndLibraries.py

from collections import defaultdict

def dfs(visited_dfs, graph_dfs, node, l_c):
    # This is the usual dfs algorithm
    # just we need to keep count of the length of our component  (l_c)
    if node not in visited_dfs:
        l_c += 1
        visited_dfs.add(node)
        for neighbour in graph_dfs[node]:
            l_c = dfs(visited_dfs, graph_dfs, neighbour, l_c)
    return l_c


# Complete the roadsAndLibraries function below.
def roadsAndLibraries(n, c_lib, c_road, a):

    if c_road >= c_lib:
        return n * c_lib

    g = defaultdict(list)
    nodes = set()
    #convert the given matrix to a graph (adjacent list)
    for pair in a:
        g[pair[0]].append(pair[1])
        g[pair[1]].append(pair[0])
        nodes.add(pair[0])
        nodes.add(pair[1])

    visited = set()
    cost = 0
    # for each unvisited node, we should find 
    # the length of the component it belongs to
    # and the calculate the cost.
    for node in nodes:
        if node not in visited:
            l_c = dfs(visited, g, node, 0)
            cost += c_lib + ((l_c - 1) * c_road)

    # in case there're cities who weren't added to the graph
    # because they had no adjecent cities (they are not present
    # in the matrix given), we should add their cost also
    isolated_cities = n - len(visited)
    cost += isolated_cities * c_lib

    return cost