pairs, source, target): graph = defaultdict(dict) for s1, s2, rate_val1, rate_val2 in currency_pairs: graph[s1][s2] = rate_val1 graph[s2][s1] = rate_val2 def backtrack(current, seen): if current == target: return 1 product = 0
pairs, source, target): graph = defaultdict(dict) for s1, s2, rate_val1, rate_val2 in currency_pairs: graph[s1][s2] = rate_val1 graph[s2][s1] = rate_val2 def backtrack(current, seen): if current == target: return 1 product = 0
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
Related questions
Question
What is the time and space complexity of this function? It is a dfs function which goes through every possible path from node to another without cycles. Time complexity could be either O(n!) or O(2^n), or is there another answer? What is the space complexity?
def dfs(currency_pairs, source, target):
graph = defaultdict(dict)
for s1, s2, rate_val1, rate_val2 in currency_pairs:
graph[s1][s2] = rate_val1
graph[s2][s1] = rate_val2
def backtrack(current, seen):
if current == target:
return 1
product = 0
if currentingraph:
neighbors = graph[current]
for neighbor in neighbors:
if neighbor not in seen:
seen.add(neighbor)
product = max(product, graph[current][neighbor] * backtrack(neighbor, seen))
seen.remove(neighbor)
return product
return backtrack(source, {source})
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